1. Chap. 7: Chemical Quantities Calculate the mass of one mole of a species from its formula. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Determine the percentage composition of a substance. Determine the empirical and molecular formulas from percentage composition or from other experimental data.

2. Calculate the mass of one mole of a species from its formula. Chemists need to quantify the amount of a compound needed for a chemical reaction. How many kilograms of iron ore is needed to make a kilogram of iron? How much hydrogen and nitrogen is required to make a metric ton of ammonia? How many grams of ammonium nitrate fertilizer are needed to yield a gram of nitrogen? How many gallons of sodium hydroxide are needed to neutralize a tank of acid? How much sulfur dioxide is required to reduce a waste stream of chromic acid solution? How much oxygen do you need to burn a gallon of gasoline?

3. Calculate the mass of one mole of a species from its formula. Molecular or Formula Mass (Weight) (all formula masses round to the tenths place) The atomic mass of an element can be found on the periodic table (e.g. Na = 23.0 amu). The molecular mass or formula mass of a compound is the sum of the atomic masses of the elements in the compound. Examples: CompoundFormula Molecular Weight Calculation (amu) OxygenO 2 2(16.0) = 32.0 WaterH 2 O 2(1.0) + 16.0 = 18.0 Sodium phosphateNa 3 PO 4 3(23.0) + 31.0 + 4(16.0) = 164.0 Iron(III) nitrateFe(NO 3 ) 3 55.8 + 3(14.0) + 9(16.0) = 241.8

4. Calculate the mass of one mole of a species from its formula. Molecular or Formula Mass (Weight) (all formula masses round to the tenths place) Find the molecular mass of Cl 2 O 7 Find the formula mass of Mg(MnO 4 ) 2 Find the molecular mass of sulfur trioxide Find the formula mass of copper (II) carbonate

5. Calculate the mass of one mole of a species from its formula. Molecular or Formula Mass (Weight) (all formula masses round to the tenths place) Find the molecular mass of Cl 2 O 7 2(35.5) + 7(16.0) = 183.0 amu Find the formula mass of Mg(MnO 4 ) 2 1(24.3) + 2(54.9) + 8(16.0) = 262.1 amu Find the molecular mass of sulfur trioxide (SO 3 ) 1(32.1) + 3(16.0) = 80.1 amu Find the formula mass of copper (II) carbonate (CuCO 3 ) 1(63.5) + 1(12.0) + 3(16.0) = 123.5 amu

6. Formula Mass Practice Name the substance and determine the formula mass: 1.K 2 S 2.Ba(ClO 3 ) 2 3.KMnO 4 4.NaHSO 4 5.PbSO 4 Go To Molecular Mass Problems HO

8. Calculate the mass of one mole of a species from its formula. Moles Scientists need to measure amounts of substances in grams, not amu. 1 mole of a substance contains the number of particles in the formula mass measured in grams (gram-formula mass) 1 particle of oxygen has a mass of 16.0 amu so 1 mole of oxygen particles has a mass of 16.0 g 1 particle of water has a mass of 18.0 amu so 1 mole of water particles has a mass of 18.0 g

9. Calculate the mass of one mole of a species from its formula. Scientists need to measure amounts of substances in grams, not amu. 1 mole of a substance contains the number of particles found in the formula mass measured in grams (gram-formula mass) 1 mole of Na 3 PO 4 particles has a mass of 164.0 grams 1 mole of Fe(NO 3 ) 3 particles has a mass of 241.8 grams What is the mass in grams of 1 mole of K 2 CrO 4 particles? What is the mass in grams of 1 mole of CH 3 OH particles?

10. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Converting Moles to Grams What is the mass in grams of 1 mole of K 2 CrO 4 particles? 194.2 g What is the mass in grams of 1 mole of CH 3 OH particles? 32.0 g What would be the mass of 1.72 moles of K 2 CrO 4 ? 1.72 mol x 194.2 g/mol = 334 g What would be the mass of 0.308 mol CH 3 OH? 0.308 mol x 32.0 g/mol = 9.86 g Pg. 183: 16 and 17

11. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Converting Grams to Moles What is the mass in grams of 1 mole of K 2 CrO 4 particles? 194.2 g What is the mass in grams of 1 mole of CH 3 OH particles? 32.0 g How many moles are in 124 g of K 2 CrO 4 ? 124 g / 194.2 g/mol = 0.639 mol If you had 1.93 g of CH 3 OH, how many moles would you have? 1.93 g / 32.0 g/mol = 0.0603 mol Pg. 183: 18 and 19

12. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Converting Moles to Particles 1 mole of any substance contains 6.02 x 10 23 particles of that substance This is called Avogadro’s number (named after a scientist). Particles can be atoms, molecules, formula units, ions –How many atoms are in 1 mol of Au?6.02 X 10 23 atoms –How many molecules are in 1 mol of CO?6.02 x 10 23 molecules –How many formula units are in 1 mol of NaCl?6.02 x 10 23 f.u. –How many ions are in 1 mol of S 2- ?6.02 x 10 23 ions

13. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Converting Moles to Particles How many molecules are in 2.64 mol of CO? 2.64 x 6.02 X 10 23 = 1.59 x 10 24 molecules How many formula units are in 0.539 mol CuSO 4 ? 0.539 x 6.02 X 10 23 = 3.24 x 10 23 f.u. How many atoms are in 24.8 mol of Mn? How many molecules are in 0.00385 mol PCl 3 ?

14. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Converting Moles to Particles How many molecules are in 2.64 mol of CO? 2.64 x 6.02 X 10 23 = 1.59 x 10 24 molecules How many formula units are in 0.539 mol CuSO 4 ? 0.539 x 6.02 X 10 23 = 3.24 x 10 23 f.u. How many atoms are in 24.8 mol of Mn? 1.49 x 10 25 How many molecules are in 0.00385 mol PCl 3 ? 2.32 x 10 21

15. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Converting Particles to Moles 7.34 x 10 22 atoms of W would be how many moles? 7.34 x 10 22 / 6.02 x 10 23 = 0.122 mol How many moles in 5.04 x 10 21 f.u. Ba(NO 3 ) 3 ? 5.04 x 10 21 / 6.02 x 10 23 = 0.00837 mol Convert 3.01 x 10 24 f.u. CaS to moles How many moles in 3.01 x 10 24 molecules of NO?

16. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Converting Particles to Moles 7.34 x 10 22 atoms of W would be how many moles? 7.34 x 10 22 / 6.02 x 10 23 = 0.122 mol How many moles in 5.04 x 10 21 f.u. Ba(NO 3 ) 3 ? 5.04 x 10 21 / 6.02 x 10 23 = 0.00837 mol Convert 3.01 x 10 24 f.u. CaS to moles5.00 mol How many moles in 3.01 x 10 24 molecules of NO?5.00 mol

17. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Review Moles  Grams:mol x formula mass Grams  Moles:grams / formula mass Moles  Particles:mol x 6.02 x 10 23 Particles  Moles:particles / 6.02 x 10 23 Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

18. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Grams to Particles How many particles are in 6.91 grams BaSO 4 ? –Convert grams to moles: 6.91 / 233.4 = 0.0296 mol –Convert moles to particles : 0.0296 x 6.02 E23 = 1.78 x 10 23 f.u. Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

19. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Particles to Grams How many grams are in 7.04 x 10 23 molecules N 2 O? –Convert molecules to moles: 7.04 E23 / 6.02 E23 = 1.17 mol –Convert moles to grams : 1.17 x 44.0 = 51.5 g Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

20. Solve problems involving the relationship between the amount of a substance in moles, mass and particles. Practice How many grams are in 1.91 x 10 24 f.u. Na 2 CO 3 –336 g How many molecules are in 154 g PCl 3 ? –6.74 x 10 23 Go to Mole Practice Problems Moles Grams Particles / Formula mass / 6.02 E23 X Formula mass X 6.02 E23

21. Determine the percentage composition of a substance. Determine the % composition of sodium chloride. Find formula mass –Formula mass of NaCl is 23.0 + 35.5 = 58.5 Divide the individual masses by the formula mass and multiply by 100% –%Na = (23.0 / 58.5) x 100% = 39.3 % –% Cl = (35.5 / 58.5) x 100% = 60.7 % Determine the % composition of ammonium nitrate Formula mass of NH 4 NO 3 is 14.0 + 4(1.0) + 14.0 + 3(16.0) = 80.0 –Percent nitrogen= 2(14.0) / 80.0 * 100= 35.0 % N –Percent hydrogen = 4(1.0) / 80.0 * 100= 5.0 % H –Percent oxygen= 3(16.0) / 80.0 * 100= 60.0 % O

22. Determine the percentage composition of a substance. Determine the % of nitrogen in urea. Formula mass of (NH 2 ) 2 CO is 2(14.0) + 4(1.0) + 12.0 + 16.0 = 60.0 % N = 28.0/60.0 * 100% = 46.7% N Determine the mass of nitrogen in 85.8 g urea 85.8 g x 0.467 = 40.1 g Determine the percentage composition of Mg(ClO 3 ) 2. Determine the percentage of Fe in FeCl 3. Determine the mass of Fe found in 51.3 grams FeCl 3.

23. Determine the percentage composition of a substance. Determine the percentage composition of Mg(ClO 3 ) 2. 24.3 + 2(35.5) + 6(16.0) = 191.3 % Mg = 24.3 / 191.3 * 100% = 12.7% % Cl = 71.0 / 191.3 * 100% = 37.1% % O = 96.0 / 191.3 * 100% = 50.2% Determine the percentage of Fe in FeCl 3. 55.9 + 3(35.5) = 162.4 %Fe = 55.9 / 162.4 * 100% = 34.4% Determine the mass of Fe found in 51.3 grams FeCl 3. 51.3g x.344 = 17.6 g

24. Percent Composition Practice Pg. 189 29-30 Pg. 191 31-33 GO TO % Comp HO

25. Determine the empirical and molecular formulas from percentage composition or from other experimental data. Empirical Formula Smallest whole number ratio of atoms in a compound Molecular Formula Actual ratio of atoms in a compound Sometimes the empirical formula and molecular formula are the same Molecular Formula Empirical Formula N2O4N2O4 NO 2 CO 2 C 6 H 12 O 6 CH 2 O C3H6O3C3H6O3

26. Determine the empirical and molecular formulas from percentage composition or from other experimental data. What is the empirical formula of a compound that contains 40.0% C, 6.7% H, and 53.3% O? 1.Assume 100 g of the compound and change % to grams. 2.Convert the grams to moles 40.0 g C / 12.0 = 3.33 mol C 6.7 g H / 1.0 = 6.7 mol H 53.3 g O / 16.0 = 3.33 mol O 3. Divide each mole quantity by the smaller number of moles 3.33 mol C / 3.33 = 1 C 6.7 mol H / 3.33 = 2 H 3.33 mol O / 3.33 = 1 O 4. Empirical formula is CH 2 O

27. Determine the empirical and molecular formulas from percentage composition or from other experimental data. What is the empirical formula of a compound that contains 26.2% N, 7.53% H and 66.4% Cl? Steps to find Empirical Formula of a compound: 1.Assume 100 g of the compound and change % to grams. 2. Convert the grams to moles 3.Divide each mole quantity by the smaller number of moles 4. Empirical formula is

28. Determine the empirical and molecular formulas from percentage composition or from other experimental data. What is the empirical formula of a compound that contains 26.2% N, 7.53% H, and 66.4% Cl? Assume 100 g of the compound and change % to grams. Convert the grams to moles 26.2 g N / 14.0 = 1.87 mol N 7.53 g H / 1.01 = 7.46 mol H 66.4 g Cl / 35.5 = 1.87 mol Cl Divide each mole quantity by the smaller number of moles 1.87 mol N / 1.87 = 1 mol N 7.46 mol H / 1.87 = 4 mol H 1.87 mol Cl / 1.87 = 1 mol Cl Empirical formula is NH 4 Cl

29. Determine the empirical and molecular formulas from percentage composition or from other experimental data. What is the empirical formula of a compound that contains 1.67 g Ce and 4.54 g I? Convert the grams to moles 1.67 g Ce / 140.1 =.0119 mol Ce 4.54 g I / 126.9 =.0358 mol I Divide each mole quantity by the smaller number of moles.0119 mol Ce /.0119 = 1 Ce.0358 mol I /.0119 = 3.01 = 3 I The empirical formula is CeI 3

30. Determine the empirical and molecular formulas from percentage composition or from other experimental data. What is the empirical formula of a compound that contains 2.74 g Na, 0.120 g H, 1.43 g C and 5.71 g O?

31. Determine the empirical and molecular formulas from percentage composition or from other experimental data. What is the empirical formula of a compound that contains 2.74 g Na, 0.120 g H, 1.43 g C and 5.71 g O? Convert the grams to moles 2.74g Na / 23.0 = 0.119 mol Na1.43 g C / 12.0 = 0.119 mol C 0.120 g H / 1.0 = 0.12 mol H5.71 g O / 16.0 = 0.357 mol O Divide each mole quantity by the smaller number of moles 0.119 mol Na / 0.119 = 1 Na0.119 mol C / 0.119 = 1 C 0.12 mol H / 0.119 = 1 H0.357 mol O / 0.119 = 3 O The empirical formula is NaHCO 3

32. Empirical Formula Practice Handout out Percent Composition and Empirical Formula practice problems.

33. Determine the empirical and molecular formulas from percentage composition or from other experimental data. To find molecular formula you need the empirical formula and the formula (molar) mass A substance has an empirical formula of CH and a molar mass of 78.0. What is the molecular formula? 1.Find the mass of the empirical formula Mass of CH is 12.0 + 1.0 = 13.0 2.Divide the molar mass by the empirical formula mass 78.0 / 13.0 = 6 3. Multiply each subscript in the EF by the number (CH) 6 = C 6 H 6

34. Determine the empirical and molecular formulas from percentage composition or from other experimental data. To find molecular formula you need the empirical formula and the formula (molar) mass A substance consists of 42.9% C and 57.1% O with a molar mass of 56.0. What is the molecular formula?

35. Determine the empirical and molecular formulas from percentage composition or from other experimental data. To find molecular formula you need the empirical formula and the formula (molar) mass A substance consists of 42.9% C and 57.1% O with a molar mass of 56.0. What is the molecular formula? Find Empirical Formula: 42.9 g C / 12.0 = 3.5857.1 g / 16.0 = 3.57 EF is CO Mass of CO is 28.0. 56.0 / 28.0 = 2 so MF = C 2 O 2

36. Practice Handout Molecular Formula Practice Problems