Chapter 4 Chemical Composition

Transcription

1 Chapter 4 Chemical Composition 4.1 (a) mole; (b) Avogadro s number; (c) empirical formula; (d) solute; (e) molarity; (f) concentrated solution 4. (a) molar mass; (b) percent composition by mass; (c) solvent; (d) concentration; (e) dilute solution; (f) dilution 4.3 The percent composition of calcium in calcium carbonate is set up as: % Ca = grams Ca grams CaCO 9.88 g Ca 100% 100% = 40.0 % Ca 4.7 g CaCO The percent composition of oxygen in calcium carbonate is set up as: % O = grams O grams CaCO 1.80 g O 100% 100% = 48.0 % O 3.75 g CaCO 3 3 The percent composition of carbon in calcium carbonate is set up as: % C = grams O grams CaCO g O 100% 100% = 1.0 % C 3.75 g CaCO Mass percent is defined as the mass of the part divided by the mass of the whole times 100%. It does not matter what mass units are used, just so long as they are the same for the part and the whole. That allows the units to cancel properly. Mass of Part Mass % = 100% Mass of Whole 30.0 μg Mass %C = 100% 50.0 μg = 60.0% C 4.6 Mass percent is defined as the mass of the part divided by the mass of the whole times 100%. It does not matter what mass units are used, just so long as they are the same for the part and the whole. That allows the units to cancel properly. Mass of Part Mass % = 100% Mass of Whole 3.54 mg Mass %N = 100% 0.0 mg = 17.7% N 4.7 To calculate the actual mass of a substance in a sample, we rearrange the percent equation in the following way: Mass of Part Mass % = 100% Mass of Whole becomes Mass % Mass of Whole Mass of Part = 100% Mass of lithium = 1. g18.8% 100% = 0.6 g lithium 4.8 To calculate the actual mass of a substance in a sample, we rearrange the percent equation in the following way: Mass of Part Mass % = 100% Mass of Whole becomes Mass % Mass of Whole Mass of Part = 100% 41

2 Mass of silicon = 348 g 46.7% 100% = 163 g silicon 4.9 Chemical formulas must have whole number subscripts. (a) H S; (b) A ratio of 1.5 oxygen atoms to 1 nitrogen atom means that the number of each atom needs to be doubled to give whole-number coefficients for both atoms. Doubling the values works because it converts 1.5 to 3 (a whole number). This gives a ratio of 3 oxygen atoms to nitrogen atoms (note that both quantities are doubled). The chemical formula is N O 3. (c) A ratio of one-half calcium ion to one chloride ion means that the number of each ion needs to be doubled to give whole-number coefficients for both ions. This gives a ratio of 1 calcium ion to chloride ions. The chemical formula for calcium chloride is CaCl Chemical formulas must have whole number subscripts. (a) PH 3 ; (b) A ratio of.5 oxygen atoms to 1 atom means that the number of each atom needs to be doubled to give whole-number coefficients for both atoms. This works because it converts.5 to 5 (whole number). This gives a ratio of 5 oxygen atoms to nitrogen atoms. The chemical formula is N O 5. (c) A ratio of 1/3 aluminum ion to one chlorate ion implies that the number of each ion must be tripled to give whole-number coefficients for both atoms. This gives a ration of one aluminum ion to three chlorate ions. The chemical formula is Al(ClO 3 ) The formulas are derived by counting the atoms of each element in the figures. (a) H SO 4 ; (b) SCl 4 ; (c) C H The formulas are derived by counting the atoms of each element in the figures. (a) H SO 3 ; (b) SCl 6 ; (c) C H Each molecule pictured contains 1 carbon atom and oxygen atoms. The formula is CO Each molecule pictured shows nitrogen and four hydrogen atoms. The formula is N H (a) The formula unit for an ionic compound is described by its formula, which shows the ratio of ions in lowest possible whole numbers. Since the formula of sodium chloride is NaCl, one formula unit is NaCl. From the image, you can see a one-to-one correspondence of sodium ions and chloride ions in the nearest atoms (you should be able to count 14 chloride and 14 sodium ions). This also gives the formula NaCl (1:1 ratio of Na + and Cl ). (b) For molecular compounds consisting of discrete molecules, the formula unit is the same as its molecular formula. In this case there are two Cl atoms in each molecule, so the formula unit is Cl. (c) For methane, another molecular compound, each molecule contains one carbon atom and four hydrogens so the molecular formula and the formula unit is CH 4. (d) The image of silicon dioxide is more difficult to analyze because it is not composed of discrete molecules. As with ionic compounds, the formula unit is a ratio with lowest possible whole numbers. On average each silicon atom shares four oxygen atoms with its neighbors, so the formula unit is one silicon and two oxygens (½ 4 oxygens); SiO. However, as with ionic compounds and molecules the formula unit can be derived from its name, silicon dioxide (a) The formula unit for an ionic compound is described by its formula, which shows the ratio of ions in lowest possible whole numbers. Since the formula of cesium chloride is CsCl, one formula unit is CsCl. From the image, you can see a one-to-one correspondence of cesium ions and chloride ions. This also gives the formula CsCl (1:1 ratio of Cs + and Cl ). (b) For molecular compounds consisting of discrete molecules, the formula unit is the same as its molecular formula. In this case there are two oxygen atoms in each molecule, so the formula unit is O. (c) For sulfur dioxide, there is one sulfur and two oxygens in each molecule, so the formula is SO. (d) For sodium, a metal, the formula unit is simply Na. 4

3 4.17 One mole of any object contains of those objects. One mole of NH 3 contains NH 3 molecules. One-half mole of NH 3 contains half of that: Molecules NH 3 = 0.5 mol NH NH 3 molecules 1 mole NH NH molecules Since there is one nitrogen atom for each NH 3, we can expect that there are N atoms as well. We can express that as follows: 1 N atom N Atoms = NH 3 = N atoms 1 NH 3 In a similar fashion we can see that there are three hydrogen atoms in each NH 3. 3 H atoms H atoms = NH 3 = H atoms 1 NH One mole of any object contains of those objects. One mole of SO molecules contains SO. If we have 1.75 moles then we have 1.75 times that number: 3 Molecules SO = 1.75 mol SO SO molecules 1 mole SO SO molecules Since there is one sulfur for each sulfur dioxide, we can expect that there are S atoms as well. We can express that as follows: S Atoms = SO 1 S atom = S atoms 1 SO The molecular formula also shows us that there are two oxygen atoms in each SO molecule, so there are twice as many O atoms than SO molecules: O Atoms = SO O atom 1 SO = O atoms 4.19 One mole of Cu S contains Cu S formula units. One-half mole of Cu S contains half that value: Formula units of CuS = 0.50 mol CuS CuS formula units 1 mole Cu S Cu S formula units 4.0 One mole of CuSO 4 contains CuSO 4 formula units. 1.5 moles of CuSO 4 contains 1.5 times that many formula units: Formula units of CuSO 4 = 1.50 mol CuSO CuSO 4 formula units 1 mole CuSO CuSO formula units 4.1 To calculate the number of S atoms in 0. mol of SO, calculate the number of molecules in 0. mol and then multiply by the number of S atoms in each SO molecule (1 S per molecule) as mapped below: 4 Map: N formula ratio A Moles Number of molecules Number of atoms Atoms S = 0. mol molecule mol 1S atom = 1 10 S atoms 1 molecule 43

4 4. To calculate the number of O atoms in 0. mol of SO, calculate the number of molecules in 0. mol and then multiply by the number of O atoms in each SO molecule ( O atoms per molecule) as mapped below: Map: N formula ratio A Moles Number of molecules Number of atoms Atoms O = 0. mol molecule mol O atom = 10 O atoms 1 molecule 4. To calculate the number of calcium ions in 1 mol of CaCl, calculate the number of formula units in 1 mol and then multiply by the number of calcium ions per formula unit (1 Ca + per CaCl ) as mapped below: Map: N formula ratio A Moles Number of formula units Number of ions Ca + ions = 1 mol formula unit mol 1 Ca = 6 10 Ca + ions 1 formula unit 4.4 To calculate the number of chloride ions in mol of CaCl, calculate the number of formula units in mol and then multiply by the number of chloride ions per formula unit ( Cl per CaCl ) as mapped below: Map: N formula ratio A Moles Number of formula units Number of ions Cl ions = mol formula unit mol Cl = 10 4 Cl ions 1 formula unit 4.5 The molar mass is the sum of the masses of the component elements in the chemical formula. The masses for one mole of each element are found on the periodic table. NaCl Mass of 1 mol of Na = 1 mol.99 g/mol =.99 g Mass of 1 mol of Cl = 1 mol g/mol = g Mass of 1 mol of NaCl = g The molar mass is g/mol Cl Mass of mol of Cl = mol g/mol = g The molar mass is g/mol CH 4 Mass of 1 mol of C = 1 mol 1.01 g/mol = 1.01 g Mass of 4 mol of H = 4 mol g/mol = 4.03 g Mass of 1 mol of CH 4 = g The molar mass is g/mol SiO Mass of 1 mol of Si= 1 mol 8.09 g/mol = 8.09 g Mass of mol of O= mol g/mol = 3.00 g Mass of 1 mol of SiO = g The molar mass is g/mol 44

5 4.6 The molar mass is the sum of the masses of the component elements in the chemical formula. The masses for one mole of each element are found on the periodic table. CsCl Mass of 1 mol of Cs = 1 mol 13.9 g/mol = 13.9 g Mass of 1 mol of Cl = 1 mol g/mol = g Mass of 1 mol of CsCl = g The molar mass is g/mol O Mass of mol of O= mol g/mol = 3.00 g The molar mass is 3.00 g/mol SO Mass of 1 mol of S = 1 mol 3.07 g/mol = 3.07 g Mass of mol of O = mol 3.00 g/mol = 3.00 g Mass of 1 mol of SO = g The molar mass is g/mol Na Mass of 1 mol of Na = 1 mol.99 g/mol =.99 g The molar mass is.99 g/mol 4.7 The molar mass is the sum of the masses of the component elements in the chemical formula. The masses for one mole of each element are found on the periodic table. (a) Hg Cl Mass of mol of Hg = mol 00.6 g/mol = 401. g Mass of mol of Cl = mol g/mol = g Mass of 1 mol of Hg Cl 47.1 g Molar mass = 47.1 g/mol (b) CaSO 4 H O Note that the mass of 1 water molecule was calculated separately as g/mol. Mass of 1 mol of Ca = 1 mol g/mol = g Mass of 1 mol of S = 1 mol 3.06 g/mol = 3.06 g Mass of 4 mol of O = 4 mol g/mol = g Mass of mol of H O = mol g/mol = g Mass of 1 mol of CaSO 4 H O = g Molar mass = g/mol (c) Cl O 5 Mass of mol of Cl = mol g/mol = g Mass of 5 mol of O = 5 mol g/mol = g Mass of 1 mol of Cl O g Molar mass = g/mol (d) NaHSO 4 Mass of 1 mol of Na = 1 mol.99 g/mol =.99 g Mass of 1 mol of H = 1 mol g/mol = g Mass of 1 mol of S = 1 mol 3.06 g/mol = 3.06 g Mass of 4 mol of O = 4 mol g/mol = g Mass of 1 mol of NaHSO 4 = g Molar mass = g/mol 45

6 4.8 The molar mass is the sum of the masses of the component elements in the chemical formula. The masses for one mole of each element are found on the periodic table. (a) K SO 4 Mass of mol of K = mol g/mol = 78.0 g Mass of 1 mol of S = 1 mol 3.06 g/mol = 3.06 g Mass of 4 mol of O = 4 mol g/mol = g Mass of 1 mol of K SO g Molar mass = g/mol (b) NiCl 6H O; Note that the mass of 1 water molecule was calculated separately as g/mol (four significant figures). Mass of 1 mol of Ni = 1 mol g/mol = g Mass of mol of Cl = mol g/mol = g Mass of 6 mol of H O = 6 mol g/mol = g Mass of 1 mol of NiCl 6H O = 7.69 g Molar mass = 7.69 g/mol (c) C H 4 Cl Mass of mol of C = mol 1.01 g/mol = 4.0 g Mass of 4 mol of H = 4 mol g/mol = 4.03 g Mass of mol of Cl = mol g/mol = g Mass of 1 mol of C H 4 Cl g Molar mass = g/mol (d) Mg(NO 3 ) Mass of 1 mol of Mg = 1 mol 4.31 g/mol = 4.31 g Mass of mol of N = 1 mol g/mol = 8.0 g Mass of 6 mol of O = 6 mol g/mol = g Mass of 1 mol of Mg(NO 3 ) = g Molar mass = g/mol 4.9 The molar mass is the sum of the masses of the component elements in the chemical formula. The masses for one mole of each element are found on the periodic table. (a) I Mass of mol of I = mol 16.9 g/mol = 53.8 g Molar mass = 53.8 g/mol (b) CrCl 3 Mass of 1 mol of Cr = 1 mol 5.00 g/mol = 5.00 g Mass of 3 mol of Cl = 3 mol g/mol = g Mass of 1 mol of CrCl 3 = g Molar mass = g/mol (c) C 4 H 8 Mass of 4 mol of C = 4 mol 1.01 g/mol = g Mass of 8 mol of H = 8 mol g/mol = g Mass of 1 mol of C 4 H g Molar mass = g/mol 4.30 The molar mass is the sum of the masses of the component elements in the chemical formula. The masses for one mole of each element are found on the periodic table. 46

7 (a) P 4 Mass of 4 mol of P = 4 mol g/mol = 1.9 g Molar mass = 1.9 g/mol (b) CrO Cl Mass of 1 mol of Cr = 1 mol 5.00 g/mol = 5.00 g Mass of mol of O = mol g/mol = 3.00 g Mass of mol of Cl = mol g/mol = g Mass of 1 mol of CrO Cl = g Molar mass = g/mol (c) CaF Mass of 1 mol of Ca = 1 mol g/mol = g Mass of mol of F = mol g/mol = g Mass of 1 mol of CaF g Molar mass = g/mol 4.31 To measure out a useful number of atoms by counting would not be possible because atoms are too small for us to see and manipulate. If you could count 1000 water molecules per second for slightly over 0 billion years, you would have counted the molecules in one small drop of water (about 0.00 ml). Instead, because we know the mass of one mole of any substance, we can carefully weigh out an amount of any substance and determine the number of atoms our sample contains. 4.3 One mole is of any object. Scientists use the mole because atoms, ions, and molecules are too small to work with individually. Large collections of these units are more reasonable to deal with The molar mass and atomic mass have the same numerical value but different units. If the molar mass of LiCl is 4.39 g/mol, the mass of one formula unit of LiCl is 4.39 amu The molar mass and atomic mass have the same numerical value but different units. If the mass of one formula unit of CuCl is amu, then the mass of 1.00 mol of CuCl is g (i.e. the molar mass is g/mol) The units for molar mass are grams per mole: g Molar mass = mol This means that if you have the numbers of grams and moles, you can calculate the molar mass by dividing the number of grams by the number of moles of substance this mass represents. In this case, you are given grams (1.0) and molecules of substance. NA Map: Number of molecules Moles Molecules = molecules mole molecules = mol Molar mass = 1.0 g mol = 35.9 g/mol 4.36 In this problem you are asked to convert from molecules to mass. You can make this conversion by first converting from molecules to moles; then use the molar mass to calculate mass of the sample. Map: N A Number of molecules Moles Mass in grams 47

8 Mass = molecules mole g = 49.0 g molecules mole 4.37 The conversion from mass to number of moles uses the following problem solving map ( is the molar mass): Map: Mass in grams Moles (a) The molar mass of KHCO 3 is g/mol (calculated by adding the masses of the component elements of the chemical formula). Moles KHCO 3 = 10.0 g KHCO3 1 mol KHCO3 = mol KHCO g KHCO 3 (b) The molar mass of H S is g/mol. 1 mol HS Moles H S = 10.0 g HS = 0.93 mol H S g H S (c) The molar mass of Se is mol Se Moles Se = 10.0 g Se = 0.17 mol Se g Se (d) The molar mass of MgSO 4 is g/mol. 1 mol MgSO4 Moles MgSO 4 = 10.0 g MgSO4 = mol MgSO g MgSO The conversion from mass to number of moles uses the following problem solving map ( is the molar mass): Map: Mass in grams Moles (a) The molar mass of SO is g/mol. 1 mol SO Moles SO = g SO = mol SO g SO (b) The molar mass of Na SO 4 is g/mol. 1 mol Na SO4 Moles Na SO 4 = g Na SO4 = mol Na SO g Na SO (c) The molar mass of BaSO 4 is 3.4 g/mol. 4 1 mol BaSO4 Moles BaSO 4 = g BaSO4 = mol BaSO g BaSO (d) The molar mass of KAl(SO 4 ) is 58. g/mol. 4 48

9 1 mol KAl(SO 4) Moles KAl(SO 4 ) = g KAl(SO 4) = mol KAl(SO 4 ) 58. g KAl(SO ) 4.39 The conversion from mass to number of moles uses the following problem solving map ( is the molar mass). In these problems, if the mass is not given in grams, it must first be converted using an appropriate problem solving map: Map: Mass in grams Moles (a) The molar mass of NaCl is g/mol. 1 mol NaCl Moles NaCl = 3.5 g NaCl = mol NaCl g NaCl (b) The molar mass of C 9 H 8 O 4 is g/mol. The mass is converted to grams using 1 mg = 10 3 g. 4 Moles C 9 H 8 O 4 = 50.0 mg C9H8O g C H O mg C9H8O4 1 mol C9H8O4 = mol C 9 H 8 O g C H O (c) The molar mass of CaCO 3 is g/mol. The mass is converted to grams using 1 kg = 10 3 g. Moles CaCO 3 = 73.4 kg CaCO g CaCO 3 1 kg CaCO 3 1 mol CaCO3 = 733 mol CaCO g CaCO 3 (d) The molar mass of CuS is g/mol. The mass is converted to grams using 1 g = 10 6 g g CuS Moles CuS = 5.47 μg CuS 1 μg CuS 1 mol CuS = mol CuS g CuS 4.40 The conversion from mass to number of moles uses the following problem solving map ( is the molar mass). In these problems, if the mass is not given in grams, it must first be converted using an appropriate problem solving map: Map: Mass in grams Moles (a) The molar mass of K SO 4 is g/mol. 1 mol KSO 4 Moles K SO 4 = 7. g KSO 4 = mol K SO g K SO 4 (b) The molar mass of C 18 H 1 NO 4 is g/mol. The mass is converted to grams using 1 mg = 10 3 g. Moles C 18 H 1 NO 4 = mg C18 H1NO g C H NO mg C18H 1NO4 1 mol C H NO g C H NO = mol C 18 H 1 NO 4 (c) The molar mass of Fe 3 O 4 is 1.55 g/mol. The mass is converted to grams using 1 kg = 10 3 g. 49

10 Moles Fe 3 O 4 =.8 kg Fe3O g Fe O kg Fe3O4 1 mol Fe3O4 = 1. mol Fe 3 O g Fe O 3 4 (d) The molar mass of C 1 H O 11 is g/mol. The mass is converted to grams using 1 g = 10 6 g. Moles C 1 H O 11 = 5.00 μg C1 HO g C H O μg C1 HO11 1 mol C H O g C H O 1 11 = mol C 1 H O If the molar mass of a substance is relatively small, it will take more moles of that substance to equal 1 gram than it would take of a substance with a larger molar mass. This means that carbon, which has the smallest molar mass of the substances given, contains the most moles of atoms in a 1.0-g sample. 4.4 If the molar mass of a substance is relatively large, it will take fewer moles of that substance to equal 5.0 g than it would take of a substance with a smaller molar mass. This means that silver, which has the highest molar mass of the substances given, contains the least moles of atoms in a 5.0-g sample To convert moles to grams we use the following problem solving map ( = molar mass) Map: Moles Mass in grams (a) The molar mass of Ba(OH) is g/mol g BaSO 4 Mass Ba(OH) =.50 mol BaSO 4 = 48 g Ba(OH) 1 mol BaSO (b) The molar mass of Cl is g/mol g Cl Mass Cl =.50 mol Cl = 177 g Cl 1 mol Cl (c) The molar mass of K SO 4 is g/mol g KSO 4 Mass K SO 4 =.50 mol KSO 4 = 436 g K SO 4 1 mol K SO 4 (c) The molar mass of PF 3 is g/mol g PF 3 Mass K SO 4 =.50 mol PF 3 = 0. g K SO 4 1 mol PF 4.44 To convert moles to grams we use the following problem solving map ( = molar mass) Map: Moles Mass in grams (a) The molar mass of I is g/mol g I Mass I = mol I = 190. g I 1 mol I (b) The molar mass of Mg(NO 3 ) is g/mol g Mg NO 3 Mass Mg(NO 3 ) = mol Mg NO 3 = 111 g Mg(NO 3 ) 1 mol Mg NO (c) The molar mass of SiO is g/mol g SiO Mass SiO = mol SiO = 45.1 g SiO 1 mol SiO

11 (c) The molar mass of Na 3 PO 4 is g/mol g Na3PO 4 Mass Na 3 PO 4 = mol Na3PO 4 = 1 g Na 3 PO 4 1 mol Na PO The conversion of.7 moles of Zn(CH 3 CO ) to an equivalent number of grams requires the molar mass (). We use the following problem solving map: Map: Moles Mass in grams The molar mass of Zn(CH 3 CO ) is g. Grams Zn(CH 3 CO ) =.7 mol Zn(CH3CO ) g Zn(CH3CO ) = g Zn(CH 3 CO ) mol Zn(CH CO ) The conversion of 3.4 moles of Cu(HCO 3 ) to an equivalent number of grams requires the molar mass (). We use the following problem solving map: Map: Moles Mass in grams The molar mass of Cu(HCO 3 ) is g. Grams Cu(HCO 3 ) = 3.4 mol Cu(HCO 3) g Cu(HCO 3) = g Cu(HCO 3 ) mol Cu(HCO ) (a) The conversion of 30.0 g NH 3 to an equivalent number of moles requires the molar mass (). We use the following problem solving map: Map: Mass in grams Moles The molar mass of NH 3 is g/mol. Moles NH 3 = 30.0 g NH3 mol NH3 = 1.76 mol NH g NH 3 (b) To calculate the number of NH 3 molecules in the sample, we use Avogadro s number (N A = ) and the following problem solving map: Map: Moles N A Number of molecules Molecules NH 3 = mol NH molecules = molecules NH 3 mol NH 3 (c) To calculate the number of nitrogen atoms in the sample, we look at the ratio of the elements in the chemical formula. The ratio for nitrogen in ammonia is 1 atom N/1 molecule NH 3. Therefore, the number of N atoms is the same as the number of NH 3 molecules, N atoms. (d) From the chemical formula, we know that there are 3 moles of H for each mole of NH 3. This is the formula ratio needed to calculate the moles of H in the sample of NH 3 : formula ratio Map: Moles NH3 Moles H 411

12 Moles H = mol NH3 3 mol H = 5.8 mol H 1 mol NH (a) The conversion of 15.0 g Na CO 3 to an equivalent number of moles requires the molar mass (). We use the following problem solving map: Map: Mass in grams Moles The molar mass of Na CO 3 is g/mol. Moles Na CO 3 = 15.0 g Na CO3 mol Na CO3 = 0.14 mol Na CO g Na CO 3 (b) To calculate the number of formula units in the sample, we use Avogadro s number (N A = ) and the following problem solving map: Map: Moles N A Number of formula units Formula units Na CO 3 = mol Na CO formula units mol Na CO 3 = formula units Na CO 3 (c) To calculate the number of sodium ions in the sample, we look at the ratio of the elements in the chemical formula. The ratio for sodium in Na CO 3 is Na + ions/1 formula unit. Map: formula ratio Number of formula units Number of Na ions Na + ions = formula units Na CO 3 Na ions = Na + ions formula units Na CO 3 (d) To calculate the number of moles carbonate ions in the sample, we look at the ratio of the carbonate ions to formula units of Na CO 3. This ratio is 1 mol CO 3 per mol of Na CO 3. Therefore, the number of carbonate ions is the same as the number of moles of Na CO 3 : 0.14 mol CO The number of atoms per mole of substance depends on the number of atoms in the chemical formula. H SO 4 has the most atoms per mole because it has more atoms per molecule. Na has the least atoms per mole The number of atoms per mole of substance depends on the number of atoms in the chemical formula. C H 6 has the most atoms per mole of substance, and Fe has the least atoms per mole of substance To calculate the number of molecules in g of water, we need to find the appropriate conversion factors. This can be done using the following conversion map: Map: Mass in grams Moles A Number of molecules N Note that the units of molar mass (g/mol) serve as a connection between mass and moles, and the units of Avogadro s number (molecules/mole) serve as a connection between molecules and moles. The molar mass of water is 18.0 g/mol. 41

13 Molecules H O = g HO mol HO 18.0 g H O molecules HO mol H O = molecules H O 4.5 To calculate the number of SiO formula units in 1 grain of sand ( g) we need to find the appropriate conversion factors. Assuming that sand is 100% SiO, we use the following conversion map: Map: Mass in grams Moles A Number of formula units N Note that the units of molar mass (g/mol) serve as a connection between mass and moles, and the units of Avogadro s number (formula units/mole) serve as a connection between formula units and moles. The molar mass of SiO is g/mol. Formula units SiO = g SiO mol SiO g SiO formula units SiO mol SiO = formula units SiO 4.53 To calculate the number of formula units we use the following conversion map: N Map: Mass in grams Moles A Number of formula units (a) Br = g/mol mol Br formula units Br Formula units Br = 50.0 g Br g Br mol Br = formula units Br (b) MgCl = 95.1 g/mol Formula units MgCl = 50.0 g MgCl mol MgCl formula units MgCl g MgCl mol MgCl = formula units MgCl (c) H O = 18.0 g/mol Formula units H O = 50.0 g HO = formula units H O mol HO 18.0 g H O formula units HO mol H O (d) Fe = g/mol mol Fe formula units Fe Formula units Fe = 50.0 g Fe g Fe mol Fe = formula units Fe 4.54 To calculate the number of formula units we use the following conversion map: N Map: Mass in grams Moles A Number of formula units (a) Cu = g/mol 413

14 mol Cu formula units Cu Formula units Cu = g Cu g Cu mol Cu = formula units Cu (b) NaBr = g/mol mol NaBr formula units NaBr Formula units NaBr = g NaBr g NaBr mol NaBr = formula units NaBr (c) SO = g/mol Formula units SO = g SO mol SO g SO formula units SO mol SO = formula units SO (d) NH 4 Cl = g/mol Formula units NH 4 Cl = g NH4Cl = formula units NH 4 Cl mol NH4Cl g NH4Cl formula units NH4Cl mol NH Cl 4.55 To calculate the number of atoms or ions of each element in g of each substance, the best strategy is to first calculate the number of formula units. Once you have determined this, use the formula ratios to calculate the number of atoms or ions. In part (b), for example, you will calculate the formula units of Ca(NO 3 ) and then use formula ratios to calculate the number of Ca + and NO 3 ions. Two problem solving maps are applied: Number of formula units: N Mass in grams Moles A Number of formula units Number of ions or atoms: formula ratio Number of formula units ions or atoms (a) Since there we are only looking for atoms of one element, we can combine the two problem solving maps. The appropriate conversions are: H =.016 g/mol; H atom = 1 H molecule Atoms = g H (b) Ca(NO 3 ) = g/mol mol H.016 g H H mol H H H atom = H atoms 4 Formula units Ca(NO 3 ) = g Ca NO 3 mol Ca NO Ca NO Ca NO3 = Ca(NO 3 ) mol Ca NO Ca + ions: 1 Ca + ion = 1 Ca(NO 3 ) Ca + ions = Ca NO 3 1 Ca 1Ca NO 3 = Ca + ions NO 3 ions: NO 3 ions = 1 Ca(NO 3 ) 414

15 NO 3 ions = Ca NO 3 NO 3 1Ca NO 3 = NO 3 ions (c) N O = 60.0 g/mol Formula units N O = g NO 4 N atoms = N O 4 O atoms = N O N 1N O O 1N O mol NO 60.0 g NO = N atoms = O atoms NO = N O mol N O (d) K SO 4 = g/mol Formula units = g KSO4 mol KSO g K SO KSO4 = K SO 4 mol K SO 4 K + ions: K + ions = 1 K SO 4 K + ions = K SO 4 K = K + ions 1K SO 4 SO 4 ions: 1 SO 4 ion = 1 K SO 4 SO 1SO4 4 ions = KSO4 = SO 4 ions 1K SO To calculate the number of atoms or ions of each element in g of each substance, the best strategy is to first calculate the number of formula units. Once you have determined this, use the formula ratios to calculate the number of atoms or ions. In part (a), for example, you will calculate the formula units of BaSO 4 and then use formula ratios to calculate the number of Ba + and SO 4 ions. Two problem solving maps are applied: Number of formula units: N Mass in grams Moles A Number of formula units Number of ions or atoms: formula ratio Number of formula units ions or atoms (a) BaSO 4 = 3.36g/mol Formula units = g BaSO4 Ba + ions = SO 4 ions = BaSO BaSO 4 mol BaSO g BaSO 4 1 Ba = Ba + ions 1BaSO 1SO 4 4 1BaSO BaSO4 = BaSO 4 mol BaSO = SO 4 ions Total ions = ( Ba + ions) + ( SO 4 ions) = ions You can also calulate the number of sulfur and oxygen atoms: 4 415

16 S atoms = O atoms = BaSO BaSO 4 1S = S atoms 1BaSO 4 4 O = O atoms 1BaSO 4 (b) Mg 3 (PO 4 ) = 6.87 g/mol; Formula units = g Mg3PO4 1mol Mg PO 6.87 g Mg PO Mg PO mol Mg PO = Mg 3 (PO 4 ) 3 Mg 3 4 = Mg + ions 1Mg PO Mg + ions = Mg PO 3 4 PO 4 3 ions = Mg (PO ) 3 4 PO 3 4 1Mg (PO ) 3 4 = PO 4 3 ions Total ions = ( Mg + ions) + ( PO 4 3 ions) = ions You can also calulate the number of phosphorus and oxygen atoms: P atoms = P Mg3PO4 = P atoms 1Mg PO O atoms = Mg PO O = O atoms 1Mg PO 3 4 (c) Since we are only looking for one type of atom, we can combine the two problem solving maps. The appropriate conversions are: O = 3.00 g/mol; O atoms = 1 O formula unit O atoms = g O mol O 3.00 g O O mol O O O atom = O atoms (d) KBr = g/mol; mol KBr KBr Formula units = g KBr = KBr g KBr mol KBr K + ions = Br ions = KBr KBr 1K = K + ions 1KBr 1 Br = Br ions 1KBr Total ions = ( K + ions + ( Br ions) = ions 4.57 To calculate the mass of molecules of SO we use the following conversion map: Map: N Number of molecules A Moles Mass in grams The molar mass of SO is g/mol. 416

17 Mass SO = molecules SO 1 mol SO g SO = 6.8 g SO moleculesso 1 mol SO 4.58 To calculate the mass of molecules of H SO 4 we use the following conversion map: Map: N Number of molecules A Moles Mass in grams The molar mass of H SO 4 is g/mol. Mass H SO 4 = molecules H SO 1 = 0.9 g H SO mol HSO molecules H SO g H SO 1 mol H SO The substance that has the most nitrogen atoms per 5.0-g sample will also have the highest number of moles of nitrogen atoms. We use the following conversion map: Map: formula ratio Mass in grams Moles Moles of N atoms The molar mass of NH 3 is g/mol and there is 1 mol N atoms per mol NH 3. Moles N atoms = 5.0 g NH3 mol NH3 1 mol N atoms = 1.47 mol N atoms g NH mol NH3 3 The molar mass of NH 4 Cl is g/mol and there is 1 mol N atoms per mole NH 4 Cl. Moles N atoms = 5.0 g NH4Cl mol NH Cl 1 mol N atoms = mol N atoms g NH Cl mol NH4Cl 4 The molar mass of NO is g/mol and there is 1 mol N atoms per mole NO. 4 Moles N atoms = 5.0 g NO mol NO 1 mol N atoms = mol N atoms g NO mol NO The molar mass of N O 3 is 76.0 g/mol and there are mol N atoms per mole N O 3. Moles N atoms = 5.0 g NO3 mol NO 3 mol N atoms = mol N atoms 76.0 g N O mol NO3 3 Because NH 3 has the highest number of moles of nitrogen atoms, we know that it has the most nitrogen atoms per 5.0-g sample The substance that has the most chlorine atoms per g sample will also have the highest number of moles of chlorine atoms. We use the following conversion map: Map: formula ratio Mass in grams Moles Moles of Cl atoms The molar mass of NaCl is g/mol and there is 1 mol Cl atoms per mol NaCl. 417

18 mol NaCl Moles Cl atoms = g NaCl g NaCl 1 mol Cl atoms mol NaCl = mol Cl atoms The molar mass of PCl 3 is g/mol and there are 3 mol Cl atoms per mole PCl 3. Moles Cl atoms = g PCl3 mol PCl3 3 mol Cl atoms =.185 mol Cl atoms g PCl mol PCl3 3 The molar mass of CaCl is g/mol and there are mol Cl atoms per mole CaCl. Moles Cl atoms = g CaCl mol CaCl mol Cl atoms = 1.80 mol Cl atoms g CaCl mol CaCl The molar mass of HClO is g/mol and there is 1 mol Cl atoms per mole HClO. Moles Cl atoms = g HClO mol HClO 1 mol Cl atoms = mol Cl atoms g HClO mol HClO Because PCl 3 has the highest number of moles of chlorine atoms, we know that it has the most chlorine atoms per g sample No. Molecules of the same substance have the same percent compositions. 4.6 Percent composition can be used to compare the composition of different substances. Copper ore, for example, is analyzed for percent copper composition, and fertilizers are analyzed for their percent nitrogen content. Percent composition can also be used to determine the empirical formula of a compound The empirical formula shows the relative amounts of each atom in a compound, expressed as small whole numbers. The molecular formula shows the exact numbers of each atom present in one molecule of that compound Chemical formulas are much more compact and give the same information (i.e. the relative composition of substances). For example, we could report that water is 88.81% oxygen and 11.19% hydrogen by mass or simply report that its formula is H O The empirical and molecular formulas are different if the subscripts in the molecular formula are all divisible by a common factor other than 1. For example, in the formula H O, both subscripts are divisible by, so the empirical formula (HO) is different than the molecular formula. Of the substances listed, those with different empirical and molecular formulas are: Molecular Empirical H O HO N O 4 NO 4.66 The empirical and molecular formulas are the same if the subscripts in the molecular formula are not divisible by a common factor other than 1. For example, in the formula N O 3, the subscripts have no common factors other than 1, so the empirical formula is the same as the molecular formula. Of the substances listed, those with the same empirical and molecular formulas are N O 3 and NaCl The empirical and molecular formulas are the same if the subscripts in the molecular formula are not divisible by a common factor other than 1. Often, it is desirable to simplify the molecular formula in order 418

19 to determine if the molecular and empirical formulas are different. For part (d) HO CC 4 H 8 CO H can be simplified to C 6 H 10 O 4. Molecular Common Empirical Formula Factor Formula (a) P 4 O 10 P O 5 (b) Cl O 5 none same as molecular (c) PbCl 4 none same as molecular (d) C 6 H 10 O 4 C 3 H 5 O 4.68 The empirical and molecular formulas are the same if the subscripts in the molecular formula are not divisible by a common factor other than 1. Molecular Common Empirical Formula Factor Formula (a) As 4 O 6 As O 3 (b) H S HS (c) CaCl none same as molecular (d) C 3 H 6 3 CH 4.69 The empirical and molecular formulas are the same if the subscripts in the molecular formula are not divisible by a common factor other than 1. Molecular Common Empirical Formula Factor Formula (a) C 6 H 4 Cl C 3 H Cl (b) C 6 H 5 Cl none same as molecular (c) N O 5 none same as molecular 4.70 The empirical and molecular formulas are the same if the subscripts in the molecular formula are not divisible by a common factor other than 1. Molecular Common Empirical Formula Factor Formula (a) N O 4 NO (b) H C O 4 HCO * (c) C H 4 O CH O *It is easier to find the empirical formula of CH 3 CO H if the molecular formula is simplified to C H 4 O NO and N O 4 have the identical empirical formulas (NO ). The empirical formulas of the other compounds are different than NO. Molecular Common Empirical Formula Factor Formula N O none N O NO none NO NO none NO N O 3 none N O 3 N O 4 NO N O 5 none N O C H 4 and C 3 H 6 have identical empirical formulas (CH ). The empirical formulas of the other compounds are different than CH. Molecular Common Empirical Formula Factor Formula CH 4 none CH 4 C H 4 CH C 3 H 6 3 CH C 4 H 1 4 CH 3 C 6 H 6 6 CH 419

20 4.73 The empirical formula shows the whole-number ratio of moles of each element in a compound. To determine the empirical formula of a compound, we calculate the number of moles of each element in the sample, and then determine the mole ratios. When presented with percent composition data, it is easiest to assume that we have exactly 100 grams of the substance. (a) The percent composition of our sample is 7.36% Fe and 7.64%O. If we have a 100-gram sample, it will contain 7.36 g Fe and 7.64 g O. We can convert these masses to the equivalent number of moles as follows: 1 mol Fe Moles Fe = 7.36 g Fe = 1.96 mol Fe g Fe 1 mol O Moles O = 7.64 g O = 1.78 mol O g O Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol Fe). moles Fe 1.96 mol Fe 1 mol Fe moles Fe 1.96 mol Fe 1 mol Fe moles O 1.78 mol O mol O moles Fe 1.96 mol Fe 1 mol Fe Notice that one of the ratios is not a whole number. Since the subscripts of chemical formulas must be whole numbers (i.e. you can t have fractions of atoms), we multiply each ratio by 3 to convert the ratio into a whole number. We find that the empirical formula has 4 moles of oxygen for every 3 moles of iron. The empirical formula is Fe 3 O 4. (b) The percent composition of our sample is 58.53% C, 4.09% H, 11.38% N, and 5.99% O. If we have a 100 gram sample, it will contain g C, 4.09 g H, g N, and 5.99 g O. We can convert these masses to the equivalent number of moles as follows: 1 mol C Moles C = g C = mol C 1.01 g C Moles H = 4.09 g H 1 mol H = 4.06 mol H g H 1 mol N Moles N = g N = 0.81 mol N g N 1 mol O Moles O = 5.99 g O = 1.64 mol O g O Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol N). moles C mol C mol C moles N 0.81 mol N 1 mol N 40

21 moles H 4.06 mol H 5.00 mol H moles N 0.81 mol N 1 mol N moles N 0.81 mol N 1 mol N moles N 0.81 mol N 1 mol N moles O 1.64 mol O.000 mol O moles N 0.81 mol N 1 mol N The empirical formula has 6 moles of carbon, 5 moles of hydrogen, and moles of oxygen for every 1 mole of nitrogen. The empirical formula is C 6 H 5 NO The empirical formula shows the whole-number ratio of moles of each element in a compound. To determine empirical formula of a compound, we calculate the number of moles of each element in the sample and then determine the mole ratios. When presented with percent composition data, it is easiest to assume that we have exactly 100 grams of the substance. (a) The percent composition of our sample is 85.6% C and 14.38%H. If we have a 100 gram sample, it will contain 85.6 g C and g H. We can convert these masses to the equivalent number of moles as follows: 1 mol C Moles C = 85.6 g C = 7.19 mol C 1.01 g C 1 mol H Moles H = g H = 14.7 mol H g H Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol C). moles C 7.19 mol C 1 mol C moles C 7.19 mol C 1 mol C moles H 14.7 mol H.001 mol H moles C 7.19 mol C 1 mol C Rounding the mole ratios to nearest whole number we find that for each mole of carbon there are moles of hydrogen. The empirical formula is CH. (b) The percent composition of our sample is 63.15% C, 5.30% H, and 31.55% O. If we have a 100-gram sample, it will contain g C, 5.30 g H, and g O. We convert these masses to the equivalent number of moles as follows: 1 mol C Moles C = g C = 5.58 mol C 1.01 g C 1 mol H Moles H = 5.30 g H = 5.6 mol H g H 41

22 1 mol O Moles O = g O = 1.97 mol O g O Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol O). moles C 5.58 mol C.667 mol C moles O 1.97 mol O 1 mol O moles H 5.6 mol H.67 mol H moles O 1.97 mol O 1 mol O moles O 1.97 mol O 1 mol O moles O 1.97 mol O 1 mol O The ratios are not all whole numbers and must be converted to whole numbers by multiplying by an appropriate factor. In this case, the fractional portion (0.667) represents /3, so we should be able to convert these values to whole numbers by multiplying each ratio by 3. Thus for every 3 moles of oxygen we have 8 moles of carbon and 8 moles of hydrogen. The empirical formula is C 8 H 8 O The percent composition of our sample is 73.19% C, 19.49% O, and 7.37% H. If we have a 100-gram sample, it will contain g C, g O, and 7.37 g H. To determine the empirical formula, we convert the masses of each element to the equivalent number of moles, and then determine the relative number of moles of each element in the substance. 1 mol C Moles C = g C = mol C 1.01 g C 1 mol O Moles O = g O = 1.18 mol O g O 1 mol H Moles H = 7.37 g H = 7.31 mol H g H Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol O). moles C mol C mol C moles O 1.18 mol O 1 mol O moles O 1.18 mol O 1 mol O moles O 1.18 mol O 1 mol O moles H 7.31 mol H 6.00 mol H moles O 1.18 mol O 1 mol O Rounding the mole ratios to the nearest whole numbers, we find that chemical formula has 5 moles of carbon, 6 moles of hydrogen, and 1 mole of oxygen. The empirical formula for eugenol is C 5 H 6 O The percent composition of our sample is 5.66% Ca, 1.30% Si, and 35.04% O. If we have a 100 gram sample, then it will contain 5.66 g Ca, 1.30 g Si, and g O. To determine the empirical formula, we 4

23 convert the masses of each element to the equivalent number of moles, and then determine the relative number of moles of each element in the compound. Moles Ca = 5.66 g Ca 1 mol Ca = mol Ca g Ca 1 mol Si Moles Si = 1.30 g Si = mol Si 8.09 g Si 1 mol O Moles O = g O =.190 mol O g O Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol Si). moles Ca mol Ca mol Ca moles Si mol Si 1 mol Si moles Si mol Si 1 mol Si moles Si mol Si 1 mol Si moles O.19 mol O mol O moles Si mol Si 1 mol Si Rounding the mole ratios to the nearest whole number, we find that chemical formula has 3 moles of calcium and 5 moles of oxygen for every 1 mole of silicon. The empirical formula is Ca 3 SiO The percent composition of our sample is 37.01% C,.% H, 18.50% N, and 4.7% O. Assuming a 100 gram sample, it will contain g C,. g H, g N, and 4.7 g O. To determine the empirical formula, we convert the masses of each element to the equivalent number of moles, and then determine the relative number of moles of each element in the substance. 1 mol C Moles C = g C = 3.08 mol C 1.01 g C Moles H =. g H 1 mol H =.0 mol H g H 1 mol N Moles N = g N = 1.30 mol N g N 1 mol O Moles O = 4.7 g O =.64 mol O g O Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol N). moles C 3.08 mol C.335 mol C moles N 1.30 mol N 1 mol N 4

24 moles H.0 mol H mol H moles N 1.30 mol N 1 mol N moles N 1.30 mol N 1 mol N moles N 1.30 mol N 1 mol N moles O.64 mol O.00 mol O moles N 1.30 mol N 1 mol N Notice that two of the ratios are not whole numbers. The fractional portions (0.333 and 0.668) represent one-third (1/3) and two-thirds (/3). We can convert these values to whole numbers by multiplying each ratio by 3. The empirical formula has 7 moles of carbon, 5 moles of hydrogen, and 6 moles of oxygen for every 3 moles of nitrogen. The empirical formula is C 7 H 5 N 3 O The percent composition of strychnine is 75.4% C, 6.63% H, 8.38% N, and 9.57% O. A 100-gram sample will contain 75.4 g C, 6.63 g H, 8.38 g N, and 9.57 g O. To determine the empirical formula, we convert the masses of each element to the equivalent number of moles, and determine the relative number of moles of each element in the substance. 1 mol C Moles C = 75.4 g C = 6.80 mol C 1.01 g C 1 mol H Moles H = 6.63 g H = 6.58 mol H g H 1 mol N Moles N = 8.38 g N = mol N g N 1 mol O Moles O = 9.57 g O = mol O g O Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol O or mol N). moles C 6.80 mol C 10.5 mol C moles O mol O 1 mol O moles H 6.58 mol H 11.0 mol H moles O mol O 1 mol O moles N mol N 1 mol N moles O mol O 1 mol O moles O mol O 1 mol O moles O mol O 1 mol O Notice that ratio of carbon to oxygen is not a whole number. The fractional portion (0.5) represents onehalf (1/). We can convert this value to a whole number by multiplying each ratio by. The empirical 44

25 formula for strychnine has 1 moles of carbon, moles of hydrogen, moles of oxygen, and moles of nitrogen. The empirical formula is C 1 H N O Convert the percentages to moles of each element, assuming that we have a g sample: 1 mol C mol C = 45.4 g C = 3.78 mol 1.01 g C 1 mol H mol H =.70 g H g H =.698 mol 1 mol O mol O = g C = 3.41 mol g O Divide each by the smallest number: mol C 3.78 mol = 1.40 mol H.698 mol mol H.698 mol = mol H.698 mol mol O 3.41 mol = 1.01 mol H.698 mol Although we did not get whole numbers for the relative amounts of carbon and oxygen, these amounts can be converted to whole numbers by multiplying by 5: 1.40 mol C 5 = mol C = 7 mol C mol H 5 = mol H = 5 mol H 1.01 mol O 5 = mol O = 6 mol O Now that we have whole number element amounts, we can write an empirical formula: C 7 H 5 O Convert the percentages to moles of each element, assuming that we have a g sample: 1 mol P mol P = g P = mol g P 1 mol S mol S = g S =.010 mol 3.06 g S Divide by the smallest number: mol S.010 mol = mol P mol mol P mol = mol P mol Although we did not get a whole number for the relative amount of sulfur, these amounts can be converted to whole numbers by multiplying by 4: mol S 4 = mol S = 7 mol S mol P 4 = mol P = 4 mol P Now that we have whole number element amounts, we can write an empirical formula: P 4 S The percent composition by mass and the molar mass are needed to determine the molecular formula. 4.8 A molecular formula represents the exact numbers of atoms of each element in one molecule of the compound while an empirical formula only represents the whole-number molar ratios of each element in the compound The ratio of the molar mass to the mass calculated from the empirical formula gives the information necessary to determine the molecular formula. The mass of the empirical formula, CH O is: 45

26 Mass of 1 mol of C = 1 mol 1.01 g/mol = 1.01 g Mass of mol of H = mol g/mol =.016 g Mass of 1 mol of O = 1 mol g/mol = g Mass of 1 mol of CH O g The ratio of the molar mass to empirical formula mass is: Molar mass ratio = 90 g / mol g / mol = 3 Multiplying each of the subscripts in CH O by three, we obtain the molecular formula C 3 H 6 O The ratio of the molar mass to the mass calculated from the empirical formula gives the information necessary to determine the molecular formula. The mass of the empirical formula, NO is: Mass of 1 mol of N = 1 mol g/mol = g Mass of 1 mol of O = mol g/mol = 3.00 g Mass of 1 mol of NO g The ratio of the molar mass to empirical formula mass is: Molar mass ratio = 9 g / mol g / mol =.0 Multiplying each of the subscripts in NO by, we obtain the molecular formula N O The strategy for determining the molecular formula from percent composition and molar mass involves two key steps. First, determine the empirical formula and the molar mass of the empirical formula from the percent composition data. Next, use the ratio of the molar mass of the compound to the empirical formula mass to determine the molecular formula. The percent composition of our sample is 40.00% C, 6.7% H, and 53.9% O. A 100-gram sample of the compound will contain g C, 6.7 g H, and 53.9 g O. To determine the empirical formula, we convert the masses of each element to the equivalent number of moles, and then determine the relative number of moles of each element in the compound. 1 mol C Moles C = g C = mol C 1.01 g C Moles H = 6.7 g H 1 mol H = 6.67 mol H g H 1 mol O Moles O = 53.9 g O = mol O g O Next divide the number of moles of each element by the smallest of the molar amounts (i.e mol C or mol O). moles C mol C 1 mol C moles O mol O 1 mol O moles H 6.67 mol H.00 mol H moles O mol O 1 mol O 46

27 moles O mol O 1 mol O moles O mol O 1 mol O The empirical formula has moles of hydrogen for each mole of oxygen and carbon. The empirical formula is CH O. The mass of the empirical formula is calculated as: Mass of 1 mol of C = 1 mol 1.01 g/mol = 1.01 g Mass of mol of H = mol g/mol =.016 g Mass of 1 mol of O = 1 mol g/mol = g Mass of 1 mol of CH O g The ratio of the molar mass to the empirical formula molar mass is: Molar mass ratio = 180 g / mol g / mol = 6.0 Multiplying the subscripts in CH O by 6, we obtain the molecular formula C 6 H 1 O The strategy for determining the molecular formula from percent composition and molar mass involves two key steps. First, determine the empirical formula and the molar mass of the empirical formula from the percent composition data. Next, use the ratio of the molar mass of the compound to the empirical formula mass to determine the molecular formula. The percent composition of our sample is 50.7% C, 9.9% H, and 39.4% N. A 100-gram sample of the compound will contain 50.7 g C, 9.9 g H, and 39.4 g N. To determine the empirical formula, we convert the masses of each element to the equivalent numbers of moles, and then determine the relative number of moles of each element in the compound. 1 mol C Moles C = 50.7 g C = 4. mol C 1.01 g C 1 mol H Moles H = 9.9 g H = 9.8 mol H g H Moles N = 39.4 g N 1 mol N =.81 mol N g N Next divide the number of moles of each element by the smallest of the molar amounts (i.e..81 mol N). moles C 4. mol C 1.50 mol C moles N.81 mol N 1 mol N moles H 9.8 mol H 3.49 mol H moles N.81 mol N 1 mol N moles N.81 mol N 1 mol N moles N.81 mol N 1 mol N Notice that two of the ratios are not whole numbers. The fractional portions (0.5) represent one-half (1/). Therefore, we can multiply the ratios by to produce whole numbers. The empirical formula contains 3 moles carbon and 7 moles hydrogen for every moles of nitrogen. The empirical formula is C 3 H 7 N. 47