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Chemical formulas such as HClO4 can be divided into empirical formula, molecular formula, and structural formula. Chemical symbols of elements in the chemical formula represent the elements present, and subscript numbers represent mole proportions of the proceeding elements. Note that no subscript number means a subscript of 1. From a chemical point of view, an element contained in the substance is a fundamental question, and we represent the elemental composition by a chemical formula, such as H2O for water. This formula implies that the water molecules consist of 2 hydrogen, and 1 oxygen atoms. The formula H2O is also the molecular formula of water. For non-molecular substances such as table salt, we represent the composition with an empirical formula. Sodium chloride is represented by NaCl, meaning that sodium and chlorine ratio in sodium chloride is 1 to 1. Again, the subscript 1 is omitted. Since table salt is an ionic compound, the formula implies that numbers of Na+ ions, and Cl- ions are the same in the solid. The subscript numbers in an empirical formula should have no common divisor. A structural formula reflects the bonding of atoms in a molecule or ion. For example, ethanol can be represented by CH3CH2OH. This is a simple way of representing a more elaborated structure shown on your left. Molecular structures are often beautiful, but the representation is an artwork. For example, a 3-dimensional structure of cyclohexane is shown on the right. This is a chair form, and another structure has a boat form. You will learn more about it in organic chemistry. The molecular formula of benzene is C6H6, and its empirical formula is CH. You may refer to a substance by its name, and recognize it by its properties. Properties are related to the structure and the composition of the molecules. Knowing the chemical formula is a giant step towards understanding a substance. Ions are formed through a process called REDOX. Redox (shorthand for reduction-oxidation reaction) describes all chemical reactions in which atoms have their oxidation number (oxidation state) changed. The term redox comes from the two concepts of reduction and oxidation. It can be explained in simple terms: Ionic substances exist with separate anions and cations arrayed in a regular pattern so the negative and positive charges cancel. The formula for an ionic compound can be predicted using the charges on the respective ions as subscripts. The illustration shows how this works for the combination of calcium ions and chloride ions. CHECK IF IT IS IONIC OR MOLECULAR!! If it is molecular, see above. sodium chloride crystal If it is ionic, name the first element, then name the second ion. If it is a single element, then just change the ending to an -ide. If it is more than one element, it is a polyatomic ion (one you have memorized)!! Use the name you memorized. Ionic Compounds Without a Transition Metal Ionic compounds are formed when a metal gives up its electrons to a non-metal. Basically if the compound contains a metal, it is ionic. But there are different sets of rules for transition metals. A transition metal is an element with an atomic number of 21 to 30, 39 to 48 or 57 to 80. So for a compound with any other metal, apply these rules For example: AlCl3 = aluminum chloride Na2S = sodium sulfide K2O = potassium oxide MgH2 = magnesium hydride Notice how the little numbers (subscripts) do not affect the way we name them. In going backwards (from name to formula), we will have more fun. In such a case the total charge of the (+) and (-) ions in the compound has to be ZERO. Example: What is the correct formula for calcium phosphide? Here, we have to consider the common charges for calcium and phosphorus, which are +2 and -3, respectively. Calcium, if you recall, typically loses two electrons to get a noble gas electron arrangement, and phosphorus needs three more electrons. Ca+2 and P-3 One of each would create a sum of 2 + (-3) = -1. To get a sum of zero, we need three Ca+2 ions and two P-3 for a total of So the answer is Ca3P2. Other examples: potassium oxide : K+1 and O-2 give K2O aluminum bromide Al+3 and Br-1 yield AlBr3. Check if the metal in a compound is a transition metal, you need a Roman numeral. When writing the names for these compounds you need to figure the charge by working backwards. Example: FeCl2 would be iron chloride, until you noticed that iron is a transition metal. To figure the charge, first find the charge of the negative ion (chloride). It is Cl-1. Since there are 2 of these chlorides given in the formula FeCl2, the total negative charge is -2. Therefore the iron must be a +2 to offset it. The final name is iron (II) chloride. Example: PtO2, O has a -2 charge, there are two of them, so the total negative charge is -4. Therefore platinum (Pt) must be +4. The final name is platinum (IV) oxide. Example: Fe2O3, O has a -2 charge, there are three of them, so the total negative charge is -6. The total positive charge must be +6, since there are 2 irons (Fe), then each Fe must have a charge of +3. The final name is iron (III) oxide. Writing basic ionic compound formulas. Ionic compounds: usually contain a metal and a nonmetal. Use charges to write the formula. If it is a transition metal, the Roman numeral gives the charge. Notice the -ide ending. This tells you that it is a single element, not a polyatomic ion. Examples: lithium sulfide Li+1 S-2 => Li2S Before naming a compound, you have to figure out what kind of compound it is. We will consider three types: copper (II) bromide Cu+2 Br-1 => CuBr2 Before naming a compound, you have to figure out what kind of compound it is. We will consider three types: Each polyatomic ion is a complete unit, NEVER break it up or change the numbers. Use charges just like with regular ionic compounds. Most end in -ate and -ite, only a few (cyanide, hydroxide) have an -ide ending. Examples: lithium sulfate Li+1 SO4-2 => Li2SO4 When metals are bonded to polyatomic ions, which consist of two or more atoms with one overall charge, the same rules apply, but you have to learn the names and charges of common polyatomic ions. Na2CO3 = sodium carbonate. ( This is a useful chemical in purifying others; it is sometimes called washing soda) KNO3 = potassium nitrate ( This is an ingredient of gunpowder and it is also found in fertilizer.) Aluminum sulfateThis has Al+3 and SO4-2. To get a sum charge of ZERO, we need two aluminum ions and three sulfates, so the formula becomes Al2(SO4)3. copper (II) nitrite Cu+2 NO3-1 => Cu(NO3)2. Notice the ( ), use these when you have more than one polyatomic ion. When metals are bonded to polyatomic ions, which consist of two or more atoms with one overall charge, the same rules apply, but you have to learn the names and charges of common polyatomic ions. Na2CO3 = sodium carbonate. ( This is a useful chemical in purifying others; it is sometimes called washing soda) KNO3 = potassium nitrate ( This is an ingredient of gunpowder and it is also found in fertilizer.) Aluminum sulfateThis has Al+3 and SO4-2. To get a sum charge of ZERO, we need two aluminum ions and three sulfates, so the formula becomes Al2(SO4)3. Notice that when there is more than one polyatomic group, we make use of brackets. The only difference here is that we have to specify the charge of the transition metal ion by using a Roman numeral, and keep in mind that a transition metal is an element with an atomic number of 21 to 30, 39 to 48 or 57 to 80. The reason we do this is not for the sake of nostalgia for bygone Roman numerals nor to imitate movie credits. Because transition metals can assume more than one charge, we have to specify which one is involved Example: manganese(II) oxide contains Mn+2 and O-2. So we just need one of each and the formula becomes MnO. Copper(I) oxide is Cu2O. To go backwards, we need to figure out the charge of the transition metal. Example: What is the correct name of CrCl3 ? The charge of Cr is unknown = x . But chloride = (-1). The sum of the charges has top be zero, so: x +3(-1) = 0. Answer: CrCl3 = chromium (III) chloride. The reason we do this is not for the sake of nostalgia for bygone Roman numerals nor to imitate movie credits. Because transition metals can assume more than one charge, we have to specify which one is involved Example: manganese(II) oxide contains Mn+2 and O-2. So we just need one of each and the formula becomes MnO. Copper(I) oxide is Cu2O. To go backwards, we need to figure out the charge of the transition metal. Example: What is the correct name of CrCl3 ? The charge of Cr is unknown = x . But chloride = (-1). The sum of the charges has top be zero, so: Answer: CrCl3 = chromium (III) chloride. Properties of Ionic Compounds Summary of properties of ionic compounds Writing formulas for molecular compounds. Molecular compounds: nonmetals only! DO NOT USE CHARGES!!! Prefixes give the number of each element. Examples: CO2 carbon dioxide, notice the ending change. The di means two of that element, in this case oxygen. Dinitrogen monoxide = N2O, dichlorine heptoxide = Cl2O7 The prefixes are: Covalent Compounds: These are formed from non-metals that share electrons. These are also known as molecular compounds. Because there are many sharing possibilities between two non-metals, the formula cannot be guessed unless we have a naming system that reveals the number of atoms involved. For this we use a set of prefixes: The only time we drop a prefix is if the mono is to appear at the beginning of the name. Examples: CO = carbon monoxide ( note we don't say monocarbon monoxide) CO2 = carbon dioxide dinitrogen pentoxide = N2O5 phosphorus trichloride PCl3. Note that none of the above compounds contain a metal. Metals do not form covalent compounds, so we generally don't use prefixes for compounds containing metals. Note that none of the above compounds contain a metal. Metals do not form covalent compounds, so we generally don't use prefixes for compounds containing metals. Two or more non-metallic atoms unite together to form molecular compounds and that is why they are easily identifiable. In molecular compounds, sharing of electrons take place between the molecules by covalent bonding. The sharing of a pair of electrons between atoms is called covalent bonding. Molecular compounds exist in three forms in standard temperature and pressure (STP) - solid, liquid and gas. Molecular compounds are bad conductors of electricity. These compounds are not as hard in comparison with ionic compounds. In a molecular compound, the attraction between the molecules is called inter-molecular attraction. The intermolecular force of attraction is a weak force and therefore less energy is required to separate these molecules from each other. Therefore, they have low melting and boiling points. Molecular compounds which have their own positive and negative charge are called polar molecules. Some molecular compounds have no charge separation due to irregular shape and are called non-polar molecules. Polar molecules are soluble in water and insoluble in non-polar solvents like organic solvents. In accordance with the naming convention, the elements present in a molecular compound are defined based on increasing group number followed by a prefix which denotes the number of atoms present in that group. Unless the first atom is a mono, a prefix is added in all other cases. For example, the systematic name of water is di hydrogen mono oxide (H20). Examples of molecular compounds are ammonia (HN3), water (H2O), carbon-dioxide (CO2) and so on. the ratio of atoms is a compound is fixed. compounds contain 2 or more elements. compounds may be decomposed to simpler forms of matter. one mole of molecules is equal to Avogadros number of particles. ionic compounds often contain metal ions. compound containing polyatomic ions and metal are considered to be both ionic and covalent. molecular or covalent compounds contain all nonmetals. molecules may be polar or nonpolar. polar molecules are asymetrical non polar molecules are symetrical. http://www.chalkbored.com/lessons/chemistry-11/naming-review.pdf A sample of any element with a mass equal to that element's atomic weight (in grams) will contain precisely one mole of atoms (6.02 x 1023 atoms). Just as the term dozen refers to the number twelve, the mole represents the number 6.02 x 1023. (If you're confused by the form of this number refer to our The Metric System module). Now that's a big number! While a dozen eggs will make a nice omelet, a mole of eggs will fill all of the oceans on earth more than 30 million times over. The mole is used when we're talking about numbers of atoms and molecules. Atoms and molecules are very tiny things. A drop of water the size of the period at the end of this sentence would contain 10 trillion water molecules. Instead of talking about trillions and quadrillions of molecules (and more), it's much simpler to use the mole. The number of objects in one mole, that is, 6.02 x 1023, is commonly referred to as Avogadro's number. Amadeo Avogadro.'); Amadeo Avogadro was an Italian physics professor who proposed in 1811 that equal volumes of different gases at the same temperature contain equal numbers of molecules. While these early estimates have since been refined, they led to the concept of the mole - that is, the theory that in a defined mass of an element (its atomic weight) there is a precise number of atoms: Avogadro's number. For example, helium has an atomic weight of 4.00. Therefore, 4.00 grams of helium will contain one mole of helium atoms. You can also work with fractions (or multiples) of moles: Other atomic weights are listed on the periodic table (see our Periodic Table page). For each element listed, measuring out a quantity of the element equal to its atomic weight in grams will yield 6.02 x 1023 atoms of that element. The atomic weight of an element identifies both the mass of one mole of that element and the total number of protons and neutrons in an atom of that element. How can that be? Let's look at hydrogen. One mole of hydrogen atoms will weigh 1.01 grams. Each hydrogen atom consists of one proton surrounded by one electron. But remember, the electron weighs so little that it does not contribute much to an atom's weight. Ignoring the weight of hydrogen's electrons, we can say that one mole of protons (H nuclei) weighs approximately one gram. Since protons and neutrons have about the same mass, a mole of either of these particles will weigh about one gram. For example, in one mole of helium, there are two moles of protons and two moles of neutrons - four grams of particles. Practice Problems Molecular Weight: If you stand on a scale with a friend, the scale will register the combined weight of both you and your friend. When atoms form molecules, the atoms bond together, and the molecule's weight is the combined weight of all of its parts. For example, every water molecule (H2O) has two atoms of hydrogen and one atom of oxygen. One mole of water molecules will contain two moles of hydrogen and one mole of oxygen. A bottle filled with exactly 18.02 g water will contain 6.02 x 1023 water molecules. The concept of fractions and multiples described above also applies to molecules: 9.01 g of water would contain 1/2 mole, or 3.01 x 1023 molecules. You can calculate the molecular weight of any compound simply by summing the weights of atoms that make up that compound. Guidelines for solving elemental analysis problems Given the following mass percent composition, determine the empirical formula. Assume 100 g sample: 49.5 g C; 5.2 g H; 28.8 g N; 16.5 g O Divide each mass number by the molar mass: Divide each molar amount by smallest amount (4.121 mol C/1.031 mol O) = (4 mol C/mol O) (5.159 mol H/1.031 mol O) = (5 mol H/mol O) (2.056 mol N/1.031 mol O) = (2 mol N/mol O) (1.031 mol O/1.031 mol O) = (1 mol O/mol O) Empirical Formula is then C4H5N2O What is the chemical formula for this compound if the molar mass is 194.2 g/mol? Listing of Example Problems: Determining Chemical Formulas Remember: all the mass of the original sample must be accounted for Police officers confiscate a packet of white powder, which they believe contains heroin. Purification by a forensic chemist gave a 38.7 mg sample for combustion analysis. This sample gave 97.7 mg CO2 and 20.81 mg H2O. A second sample was analyzed for its nitrogen content, which was 3.8 %. Show by calculations whether or not these data are consistent with the formula for heroin C21H23O5N. Next, divide each number above by smallest molar amount Finally we find that the empirical and chemical formula is C21H23NO5 A chemical compound was found to contain only Fe, C, and H. When 5.00 g of this compound was completely burned in O2, 9.1 g of CO2 and 1.80 g of H2O were produced. Find the percentage by mass of each element in the original compound and its empirical formula. 0.206 mol C/0.042 mol Fe = 5 mol C/mol Fe 0.198 mol H/0.042 mol Fe = 5 mol H/mol Fe 0.042 mol Fe/0.042 mol Fe = 1 mol Fe/mol Fe Online lesson: Finding empirical formulas Online lesson: Calculating empirical formulas Online lesson: Calculating empirical formulas based on percent composition Finding empirical formulas practice http://www.youtube.com/watch?v=J-Am0IpXlNU&NR=1 Video: Calculating percentage composition The percent composition (percentage composition) of a compound is a relative measure of the mass of each different element present in the compound. To calculate the percent composition (percentage composition) of a compound Calculate the molecular mass (MM): Calculate the total mass of Na present: 1 Na is present in the formula, mass = 22.99 Calculate the percent by weight of Na in NaCl: Calculate the total mass of Cl present: 1 Cl is present in the formula, mass = 35.45 Calculate the percent by weight of Cl in NaCl: The answers above are probably correct if %Na + %Cl = 100, that is, Calculate the molecular mass (MM): Calculate the total mass of Na present: 2 Na are present in the formula, mass = 2 x 22.99 = 45.98 Calculate the percent by weight of Na in Na2SO4: Calculate the total mass of S present in Na2SO4: 1 S is present in the formula, mass = 32.06 Calculate the percent by weight of S present: Calculate the total mass of O present in Na2SO4: 4 O are present in the formula, mass = 4 x 16.00 = 64.00 Calculate the percent by weight of O in Na2SO4: The answers above are probably correct if %Na + %S + %O = 100, that is, Calculate the molecular mass (MM) of (NH4)3PO4: Calculate the total mass of N present: 3 N are present, mass = 3 x 14.01 = 42.03 Calculate the percent by mass of N present in (NH4)3PO4: Calculate the total mass of H present: 12 H are present in the formula, mass = 12 x 1.008 = 12.096 Calculate the percent by mass of H present in (NH4)3PO4: Calculate the total mass of P present: 1 P is present in the formula, mass = 30.97 Calculate the percent by mass P in (NH4)3PO4: Calculate the total mass of O present: 4 O are present in the formula, mass = 4 x 16.00 = 64.00 Calculate the percent by mass of O in (NH4)3PO4: The answers above are probably correct if %N + %H + %P + %O =100, that is, Let's start with a few definitions. The empirical formula is the simplest formula for a compound. A molecular formula is the same as or a multiple of the empirical formula, and is based on the actual number of atoms of each type in the compound. For example, if the empirical formula of a compound is C3H8 , its molecular formula may be C3H8 , C6H16 , etc. An empirical formula is often calculated from elemental composition data. The weight percentage of each of the elements present in the compound is given by this elemental composition. Let's determine the empirical formula for a compound with the following elemental composition: This means in 100 g of this compound, 40.00 g will be due to carbon, 6.72 g will be due to hydrogen, and 53.29 g will be due to oxygen. Let's determine the empirical formula for a compound with the following elemental composition: This means in 100 g of this compound, 40.00 g will be due to carbon, 6.72 g will be due to hydrogen, and 53.29 g will be due to oxygen. We will need to compare these elements to each other stoichiometrically. In order to compare these quantities, they must be expressed in terms of moles. So the next task will be to convert each of these masses to moles, using their respective atomic weights: Take notice that since the composition data was given to four significant figures, the atomic weights used in the calculation were to at least four significant figures. Using fewer significant figures may actually lead to an erroneous formula. Now that the moles of each element are known, a stoichiometric comparison between the elements can be made to determine the empirical formula. This is achieved by dividing through each of the mole quantities by which ever mole quantity is the smallest number of moles. In this example, the smallest mole quantity is either the moles of carbon or moles of oxygen (3.331 mol): The ratio of C:H:O has been found to be 1:2:1, thus the empirical formula is: CH2O. Again, as a reminder, this is the simplest formula for the compound, and not necessarily the molecular formula. Suppose we know that the molecular weight of this compound is 180 g/mol. With this information, the molecular formula may be determined. The formula weight of the empirical formula is 30 g/mol. Divide the molecular weight by the empirical formula weight to find a multiple: The molecular formula is a multiple of 6 times the empirical formula: Alternatively, the empirical and molecular formula may be determined from experimental data. Suppose we have a compound containing the elements, C, H and S. A 7.96 mg sample of this compound is burned in oxygen and found to form 16.65 mg of CO2. The sulfur in 4.31 mg of the compound is converted into sulfate by a series of reactions, and precipitated as BaSO4 . The BaSO4 was found to have a mass of 11.96 mg. The molecular weight of the compound was found to be 168 g/mol. Using this data, what is the molecular formula of the compound? The strategy will be to use stoichiometry to determine the mass percent of each of the elements in the compound, then use the mass percentages to determine the empirical formula. Notice that since all the data is in milligrams, we may carry out the calculations using milli- units throughout. The only source of carbon for the CO2 formed came from the compound, thus, determine the milligrams of carbon found in 16.65 mg of CO2 : Now that the "part" of the sample due to carbon is known, one may calculate the percent carbon in the compound, using the mass the sample as the "whole": The only source of sulfur for the precipitate of BaSO4 , came from the compound, thus, determine the milligrams of sulfur in 11.96 mg of BaSO4 : Similarly, determine the percent sulfur in the compound, using the mass of sulfur as the "part" and the mass of compound as the "whole": The percentage of hydrogen may determined by difference: From the elemental composition, we may determine the empirical formula, in the same manner as used in the first example. First, assume exactly 100 g of the compound. In 100 grams of the compound, 57.1 g would be due to carbon, 38.0 g would be due to sulfur and 4.9 g would be due to hydrogen. Convert each of these masses into moles using the corresponding atomic weight for each element: Now that the moles of each element are known, the empirical formula may be determined by dividing the moles of each element by the smallest number of moles. This yields a ratio of the number of each element in the empirical formula. The ratio of C:H:S has been found to be 4:4:1, thus the empirical formula is: C4H4S. The molar mass of the empirical formula is 84 g/mol. Since the molecular weight of the actual compound is 168 g/mol, and is double the molar mass of the empirical formula, the molecular formula must be twice the empirical formula: SUMMARY AND PRACTICE Chemical formulas are either empirical or molecular. The empirical formula represents the simplest or smallest whole number ratio of atoms while the molecular (or true) formula gives the actual number of atoms of each element in one molecule. CALCULATION OF EMPIRICAL FORMULAS The empirical formula of a compound may be determined from a knowledge of the composition by weight and the atomic weights of the elements represented in the compound. Step 1: From the percentage by weight, calculate the number of moles of each element present in 100 g of the substance. Step 2: Obtain the simplest whole number ration of moles of each element. This is done by dividing the number of moles of each element in Step 1 by the smallest number of moles calculated. Example 1: A certain hydrocarbon contains 25% hydrogen and 75% carbon by weight. Calculate its empirical formula. Solution: From a table of atomic weights: H = 1.0 and C = 12.0. Set up a tabular form: Element Weight (per 100 g sample) Number of moles (per 100 g sample) Quotient obtained by dividing by smallest # moles Simplest Ratio Empirical formula is CH4 The student should note that in Step 2, dividing through by the smallest quotient does not always give a simple whole number ratio. Whole number ratios must exist because after all we have calculated relative numbers of moles and moles refer to numbers of atoms. Any ratios which are not calculated to be whole numbers must be attributed to experimental error arising from the % determinations. If this is the case, it must not be rounded off, but instead multiplied through by the smallest factor necessary to give whole numbers. The following example illustrates this point. Example 2: A compound contains 6.841 g of chromium combined with 3.158 g of oxygen. Calculate its empirical formula. Solution: Atomic weights, Cr = 52.0 and 0 = 16.0 Element Weight Number of moles Quotient obtained by dividing by smallest # moles Simplest Ratio Cr Empirical formula is Cr2O3 PERCENT COMPOSITION FROM EMPIRICAL FORMULA The percent composition by weight of a compound can be determined from its chemical formula. Example 3: Calculate the percentage composition of K2CO3 . Solution: K = 39.1, C = 12.0, and O = 16.0 Step 1: Calculate relative weight of each element and formula weight. 2 atomic weights of K = 2 x 39.1 = 78.2 amu 1 atomic weight of C = 12.0 amu 3 atomic weights of O = 3 x 16.0 = 48.0 amu Formula weight of K2CO3 = 138.2 amu Step 2: Divide the relative weight of each element by the formula weight. Multiply by 100 to convert to percent. Total % of all elements must add to 100% CALCULATION OF MOLECULAR FORMULAS If, in addition to composition data, the molecular weight is known or data by which the molecular weight can be calculated is available, then the true or molecular formula of the substance may be determined. Example 4: A certain gas contains 85.7% carbon and 14.3% hydrogen. Its molecular weight is 42. Determine its molecular formula. Element Weight (per 100 g sample) Number of moles (per 100 g sample) Quotient obtained by dividing by smallest # moles Simplest Ratio A certain phosphate of magnesium has the following composition: Mg = 21.85%, P = 27.85%, and 0 50.31%. Determine its simplest formula. Calculate the empirical formula of the compound known to contain 0.7888 g of gold combined with 0.4260 g of chlorine. How many ounces of silver can be obtained from a pound of AgNO3 ? How many grams of oxygen are there in 100.0 g of 903% pure KCO3 ? Determine the true formula of the compound having a molecular weight of approximately 92 and containing 0.36 g of carbon, 0.090 g of hydrogen, and 0.24 g of oxygen. The empirical formula of a certain compound is AB2. If it is known to contain 60% A, having an atomic weight of 30 what must be the atomic weight of B? A sample of hydrocarbon weighing 0.5000 g produced 1.650 g of CO2 when burned in pure oxygen. What is its empirical formula? Calculate the empirical formula of a mineral if chemical analysis showed the following composition: 20.0% CaO, 15.7% MgO, and 64.3% SiO2 Hydrated calcium phosphite contains 81.63% anhydrous CaHPO3. Derive its simplest formula. Calculate the simplest formulas of the compounds formed as indicated below. (a) 1.1817 g Cu combined with 1.3183 g Cl (b) 1.2432 g Pb combined with 0.1280 g O ANSWERS TO PROBLEMS AuCl3 CaMgSi3O8 CaHPO3 ·3H2O (a) CuCl2 (b) Pb3O4 (c) FeS2 Hydrates can normally be decomposed into the anhydrous salt and water by gentle heating. From the data collected the number of molecules of hydrated water will be determined per molecule of anhydrous salt. Most IONIC compounds are HYGROSCOPIC, which means they absorb moisture (water) from the air. The positively charged metal is attracted to (and therefore pulls) a water molecule’s oxygen (its electrons). The negatively charged non-metal is attracted to (and therefore pulls) a water molecule’s hydrogens (its electrons). These two forces of attraction result in the formation of hydrates. It is important to note that the water molecules are NOT actually bonded to the compound (i.e. having water attached will NOT change the formula of the compound). Instead, water molecules are loosely attached to the ionic compound. *Please Note: The number of water molecules will depend on the elements that make up the ionic compound (and their ability to attract those water molecules). It also depends on the conditions under which the element is stored. For now, you will just be expected to be able to name those compounds, not to predict the number of water molecules that are involved. As a result, this changes the formula of the compound. Example: CuSO4 • 5H2O This formula means that for every CuSO4 in this substance, there are also five water molecules. *The dot IS NOT a multiplication sign! Name: copper(II) sulfate pentahydrate. “Penta”, meaning five and “hydrate” meaning water tell us that 5 waters are attached to this compound. The number of water molecules attached must be identified, both in the writing of the chemical’s name AND when saying the name. This means that when you hear "pentahydrate," you have to know to write the dot and then the 5H2O. http://www.kentchemistry.com/Labs/hydrateVidLab.htm Rules for Working with Significant Figures: EXAMPLES: EXAMPLES Notes on Rounding What is the easiest way to determine the charge on the cation formed by a metal? Why do calcium and magnesium form a +2 cation instead of +3 ion? Why does chlorine, Cl, form a -1 anion and not a -2 anion? Predict, write and name the formula units for the compounds formed by the ion pairs listed below? Fe3+ with Cl1- Ca2+ with PO43- NH4+ with SO42- Fe2+ with Cl1- Ca2+ with HCO31- NH4+ with CO32- Zn2+ with Cl1- Ba2+ with NO31- Al3+ with HPO42- Cu2+ with F1- Ca2+ with OH1- K+ with SO32- Draw the electron dot structures for the following atoms and ions. How do you know the number of additional electron dots needed to convert an atom like sulfur (S) to an anion like sulfide, S2-? Ne Explain why all hydrogen containing compounds are not acids like HCl(aq). What happens in a solution of a strong acid like HCl(aq) that makes the mixture acidic? What happens when water and ethanol, CH3CH2OH, are mixed? What happens when NaOH(s) is dissolved in water? What ions are formed in the solution? Tell whether the solution is acidic or basic. What is the octet rule? Tell how the octet rule relates to the subshells in an atom. Why do you suppose hydrogen does not follow the octet rule? Do the atoms in row three, Na through Cl, always follow the octet rule? Does sulfur follow the octet rule when it forms the sulfide anion, S2-? Why is the charge on a chloride ion a minus one even though the chlorine atom gains an electron to form the ion? Short essay questions What are the sources of repulsions and attractions in the hydrogen molecule. Give answer in terms of nuclei and valence electrons. What is the origin of the octet rule? Is it universally followed? If any exceptions exist, what causes them? What is the origin of the octet rule? Is it universally followed? If any exceptions exist, what causes them? Give the names for the following compounds. HCl, PCl5, NO2, Cl2O5, CO, CO2, N2O4, SO3 What does the term binary compound mean? Give the names for the following compounds. HCl, PCl5, NO2, Cl2O5, CO, CO2, N2O4, SO3 What does the term binary compound mean? Draw the Lewis dot formula for HNO3. Identify the central atom and any multiple bonds. What is the shape around the central atom in HNO3? give bond angles Indicate the polarity of bonds using +---> arrows . Draw the Lewis dot formula for HNO3. Identify the central atom and any multiple bonds. What is the shape around the central atom in HNO3? give bond angles Indicate the polarity of bonds using +---> arrows . Multiple choice 1. Which of the following pairs of atoms will form a single covalent bond. Na and Cl b. He and He c. Ne and He d. N and O Multiple choice Which of the following pairs of atoms will form a single covalent bond. Na and Cl He and He Ne and He N and O Which of the following molecules contains exactly 24 valence electrons? NO Which of the following pairs of atoms can form a double bond? He and He H and H N and H N and O Which of the following pairs of names and symbols is incorrect? silicon tetrafluoride, SiF nitrogen dioxide, NO2 phosphorus trichloride, PCl3 phosphorus pentachloride, PCl3 Which of the following sets of elements normally exists as diatomic molecules? nitrogen, neon, helium, oxygen nitrogen, oxygen, chlorine, fluorine helium, neon, argon sodium, calcium, zinc Which of the following molecules is the most polar? hydrogen chloride, HCl hydrogen iodide, HI hydrogen, H2 chlorine, Cl2 Which of the following molecules contains a multiple bond? methane, CH4 nitrogen, N2 hydrogen, H2 chlorine, Cl2 The smallest possible unit of a covalent compound is usually considered to be a(an) atom cation molecule polyatomic ion The systematic name for ICl3 iodate trichloride iodine trichlorine iridium trichloride triiodinechloride A double bond exists when two atoms share ______ electrons. two three four six Carbon dioxide is a linear molecule. The electronegativity of carbon is 2.5. The electronegativity of oxygen is 3.5. The bond between C and O is considered to be nonpolar polar ionic weak Water molecules are polar. This is only possible if the molecule is_________________ linear tetrahedral bent pyramidal Molar Mass Practice Worksheet Find the molar masses of the following compounds 1) NaBr 2) PbSO4 10) AgF Solutions to the Molar Mass Practice Worksheet: Solutions to the Molar Mass Practice Worksheet: Important note to students: All of the units given here are “grams per mole”, which may be abbreviated as “g/mol”, “grams/mol”, or “g . mol-1”, depending on how your teacher likes to see it written. They all mean the same thing, but it’s probably a good idea to use whatever your teacher showed you in class. Also, remember that if you don’t use units in your answer, the answer is wrong! All answers are rounded to the nearest 0.1 grams. Percent Composition Worksheet Find the percent compositions of all of the elements in the following compounds: H:___________ Percent Composition Worksheet - Solutions Find the percent compositions of all of the elements in the following compounds: Mole Calculation Practice Worksheet Answer the following questions: 1) How many moles are in 25 grams of water? 2) How many grams are in 4.5 moles of Li2O? 3) How many molecules are in 23 moles of oxygen? 4) How many moles are in 3.4 x 1023 molecules of H2SO4? 5) How many molecules are in 25 grams of NH3? 6) How many grams are in 8.2 x 1022 molecules of N2I6? Solutions for the mole calculation practice worksheet: 3) 1.38 x 1025 molecules 5) 8.85 x 1023 molecules More Mole Calculations! 1) How many grams does 0.500 moles of CuBr weigh? 2) How many molecules are there in 0.655 moles of C6H14? 3) How many moles are there in 2.35 x 1024 molecules of water? 4) How many grams does 5.60 x 1022 molecules of SiO2 weigh? 5) How many molecules are there in 21.6 grams of CH4? 1) How many grams does 0.500 moles of CuBr weigh? 31.8 grams 2) How many molecules are there in 0.655 moles of C6H14? 3.94 x 1023 molecules 3) How many moles are there in 2.35 x 1024 molecules of water? 3.90 moles 4) How many grams does 5.60 x 1022 molecules of SiO2 weigh? 5.59 grams 5) How many molecules are there in 21.6 grams of CH4? 8.13 x 1023 molecules Calculating Empirical and Molecular Formulas Determine the Molar mass of the following compounds BaSO4 lithium carbonate Calculate the Percent Composition of oxygen in the following compounds CH3COOH Ammonium Sulfate Calculate the empirical formula for the following compounds from percent composition 11.66 g iron, 5.01 g oxygen 40.0 percent C, 6.7 percent H, and 53.3 percent O by mass Calculate the empirical formula for the following (show your work!) (No work, No credit) 15.8% carbon and 84.2% sulfur 43.6% phosphorus and 56.4% oxygen NEED HELP? CLICK BELOW: Calculate the molecular formula for the following (show your work!) (No work, No credit) (No work, No credit) empirical formula CH, molar mass = 78 g/mol empirical formula NO2, molar mass = 92.02 g/mol caffeine, 49.5% C, 5.15% H, 28.9% N, 16.5% O by mass, molar mass = 195 g. NEED HELP? CLICK BELOW: CLICK BELOW: Answers.
1.1 1.2 2.1 3.1 3.2 3.3 4.1 5.1 5.2 5.3 6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 8.1 8.2 9.1 9.2 10.1 11.1 12.1 13.1 13.2 13.3 13.4 14.1 14.2 14.3 14.4 15.1 15.2 16.1 17.1 17.2 18.1 18.2 19.1 19.2 19.3 20.1 20.2 21.1 21.2 22.1 23.1 24.1 25.1 25.2 26.1 27.1 28.1 29.1 30.1 31.1 32.1 33.1 34.1 35.1 36.1 37.1 38.1 39.1 40.1 41.1 41.2 42.1 43.1 44.1 45.1 46.1 47.1 48.1 49.1 50.1 51.1 52.1 52.2 53.1 54.1 54.2 55.1 56.1
Radijus atoma urana iznosi 152 pm. Kada bismo uzeli 1 mg urana i poredali atome u lanac, koliko bi kilometara bio dugačak taj lanac? Izračunajte s kojim postotkom mase sudjeluju fosfor, kisik i vodik u masi molekule fosforne kiselina (H3PO4). Organski spoj sadrži samo ugljik i vodik. 7,81 mg spoja daje 26,4 mg ugljikovog(IV) oksida i 5,40 mg vode. Relativna molekulska masa spoja je ~78. Odredite empirijsku i molekulsku formulu. S koliko molekula vode kristalizira kobaltov(II) nitrat ako 14,6 mg spoja zagrijavanjem gubi 5,40 mg vode? Cinabarit je mineral žive za koji se zna da sadrži živu i sumpor. Analizom uzorka obiveno je 30,2 mg Hg i 9,64 mg SO2. Odredite formulu cinabarita. Forenzički kemičar je analizirao bijele tablete pronañene u bočici s natpisom „aspirin“ okraj beživotnog tijela. Utvrdio je da jedna tableta sadrži 60,05% kalija, 18,44% gljika i 21,51% dušika. Druga tableta sadržavala je 60% ugljika, 4,48 % vodika i 5,53% kisika. Molarna masa je 180 g/mol. Mogu li navedeni podaci pomoći forenzičaru u utvrñivanju je li smrt nastupila slučajno, radi li se o samoubojstvu ili bojstvu? Relativna molekulska masa inzulina je 5 721. Izračunaj masu molekule inzulina. Izračunaj masu inzulina prisutnu u 1 molu inzulina. Kolika je brojnost molekula inzulina prisutna u 1 molu? U reakciji neutralizacije sumporne kiseline i natrijeve lužine, reagiralo je 20ml sumporne kiseline, množinske koncentracije 0,25 mol/dm3. Izračunaj koliki je volumen natrijeve lužine bio potreban za ovu reakciju, ako je njena masena koncentracija 5 g/dm3. Izračunaj masu natrijeva klorida u 10g morske soli, ako je maseni udio natrijeva klorida 98.8%. U kakvom su agregatnom stanju ostali sastojci morske soli? U jednoj litri vode otopljeno je 5 g modre galice.Izračunaj maseni udio modre galice u toj otopini. U koliko vode treba otopiti 4g natrijeva hidroksida kako bi se dobila otopina masenog udjela 25%? Napiši kemijsku jednadžbu gorenja etena, te izračunaj koliko je ugljikovog(IV)oksida nastalo gorenjem 25 grama etena. Relativna molekulska masa inzulina je 5 721. Izračunaj masu molekule inzulina. Izračunaj masu inzulina prisutnu u 1 molu inzulina. Kolika je brojnost molekula inzulina prisutna u 1 molu? Masa molekula joda je 1,8 grama. Kolika je množina? Koliki je broj molekula? Koliki je broj atoma joda? Množina amonijaka je 0,1 mol. Kolika je masa amonijaka i koliki je broj molekula amonijaka, a koliki je broj atoma dušika i vodika? Izračunaj masu i množinu uzorka magnezijeva oksida ako je brojnost formulskih jedinki 2,16 . 10 23. Relativna molekulska masa nikotina je 162,2. Koliki je broj molekula nikotina prisutan u 2 grama nikotina? Cink je metal prisutan u mjedi (legura bakra i cinka). Koliko je grama cinka prisutno u 0,356 mola cinka? Helij je važan plin u industriji. Upotrebljava se i za punjenje balona. Koliko je mola helija u 6,46 grama helija? Metan je najzastupljeniji plin u gradskom plinu. Kolika je množina metana ako je masa 6,07 grama? Analizom nekog spoja određeno je da sadrži 56,34% fosfora i 43,66% kisika. Odredi mpirijsku formulu spoja. Odredi empirijsku formulu magnetita, važne rude željeza, ako znaš da ruda sadrži 2,36% željeza i 27,64% kisika. Odredi empirijsku formulu uzorka koji sadrži 35,24 grama željeza, 30,34 grama umpora i 45,42 grama kisika. Odredi empirijsku formulu tvari koja se upotrebljava u izradi fotografija ako znaš da zorak sadrži 1,988 grama natrija; 2,773 grama sumpora i 2,075 grama kisika. Odredi molekulsku formulu ugljikovodika (spoj koji se sastoji samo od ugljika i vodika) koji sadrži 89,94% ugljika, a ostatak je vodik. Molarna masa spoja je 40,1 g/mol. Polietilen je sastavljen od 85,63% ugljika i 14,37% vodika. a) koliko je grama svakog elementa sadržano u 100 grama polietilena? b) koliko mola svakog elementa je sadržano u 100 grama polietilena? c) koja je empirijska formula spoja? Odredi empirijsku formulu spojeva na osnovi postotnog sastava. a) 87,73% ugljika; 12,27% vodika b) 43,645 fosfora; 56,36% kisika c) 72,03% mangana; 27,97% kisika d) 47,05% kalij; 14,45% ugljika; 38,5% kisika e) 77,26% žive; 9,25% ugljika; 1,17% vodika; 12,32% kisika Odredi empirijsku formulu spojeva na osnovi postotnog sastava. a) 37,82% ugljika; 6,35% vodika; 55,83% klora b) 54,53% ugljika; 9,15% vodika; 36,32% kisika c) 74,98% ugljika; 5,24% vodika; 19,77% fluora d) 66,42% ugljika; 5,57% vodika; 28,01% klora Nitroglicerin sadrži 15,87% ugljika; 2,22% vodika; 18,51% dušika i 63,41% kisika. Odredi empirijsku formulu spoja. Odredi empirijsku formulu spoja na temelju rezultata analize: a) 75,95 grama ugljika; 9,57 grama vodika; 224,2 grama klora b) 5,52 grama ugljika; 0,464 grama vodika; 13,1 gram fluora Odredi molekulsku formulu tvari ako je empirijska formula NO2, a molarna masa a) 92 g/mol b) 46 g/mol Odredi molekulsku formulu tvari ako je empirijska formula CH2 a molarna masa a) 84 g/mol; b) 98 g/mol; c) 168 g/mol;d) 280 g/mol U jednoj takvoj sintezi iz 100 g borovog trioksida dobiveno je 25 g elementarnog bora. Koliko je iskorištenje reakcije? Koliko je borova trioksida potrebno uzeti za sintezu 25 g elementarnog bora, ako iskorištenje kemijske reakcije iz prethodnog zadatka iznosi 75%? U reakciji sudjeluju amonijak i 1500 m3 klorovodika. Izračunaj masu amonijaka koja je reagirala, pri 10 bara, i temperaturi 40°C. NH3 + HCl -> NH4Cl Ako je u reakciji željeza s kisikom utrošeno 300 dm3 kisika, odredi koliko je željeza reagiralo, te koliko je željezne rude bilo potrebno uzeti, ako sadrži 80% željeza.