Ch. 7 Review Molar Mass Calculations Question #1 (moles  mass) What is the mass of 7.50 moles of sulfur dioxide, SO 2 ? S = 32.07 g/mol O = 16.00 g/mol.

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Ch. 7 Review Molar Mass Calculations

Question #1 (moles  mass) What is the mass of 7.50 moles of sulfur dioxide, SO 2 ? S = g/mol O = g/mol 7.50 mol SO 2 = g SO 2 1 mol SO 2 g SO 2 X

7.50 mol SO 2 = g SO 2 1 mol SO 2 g SO 2 X g SO g SO 2 SO g g x g g g = x 1

7.50 mol SO 2 = g SO 2 1 mol SO 2 g SO X g SO g SO 2 SO g g x g g g = x 1 481

Question 2: (Mass  Moles) How many moles are there in g of potassium sulfate K 2 SO 4 ? K=39.10 g/mol; S=32.07 g/mol; O=16.00 g/mol g K 2 SO 4 = mol K 2 SO 4 g K 2 SO 4 1 mol K 2 SO 4 X

993.6 g K 2 SO 4 = mol K 2 SO 4 g K 2 SO 4 1 mol K 2 SO 4 X g = g K 2 SO g x g x g g g g x

Question 3: (Moles  Molecules) How many molecules are there in 4.55 moles of nitrogen (N 2 )? 4.55 mol N 2 = molecules N 2 1 mol N 2 molecules N 2 X 1 mole = 6.02 x molecules 6.02 x x 10 24

Question 4: (Moles  molecules  atoms) How many atoms of nitrogen, N, are there in 2.18 moles of nitrogen (N 2 )? 2.18 mol N 2 = atoms N 1 mol N 2 molecules N 2 X 1 mole = 6.02 x molecules 6.02 x x molecules N 2 atoms N X 2

Question 5: (molecules  moles) How many moles are there in x molecules of carbon tetrachloride (CCl 4 )? x molecules CCl 4 = mol CCl 4 molecules CCl 4 1 mol CCl 4 X 1 mole = 6.02 x molecules 6.02 x x

Question 6: (moles  volume) What is the volume occupied by 4.20 moles of oxygen gas (O 2 ) at STP? 4.20 mol O 2 = L O 2 1 mol O 2 L O 2 X 1 mole = 22.4 L for

Question 7: (volume  moles) How many moles are there in dm 3 of argon gas (Ar) at STP? dm 3 Ar = mol Ar dm 3 Ar 1 mol Ar X & 1 mole = 22.4 dm 3 for Note: 1 L = 1 dm 3

%NX 100 = g g = Question 8: (Percent composition) g N 2 O g N 2 O 4 N2O4N2O4N2O4N2O g g x g g g = x % N %OX 100 = g g = 69.6 % O

%OX 100 = g g = Question 9: (Percent composition) 29.6 % O %F=- 29.6%100% = 70.4 % F A sample of a compound that has a mass of g is analyzed. The sample is found to be made up of oxygen and fluorine only. Given that the sample contains g of oxygen, calculate the percentage composition of the compound. Alternative method: g – g = g F %FX 100 = g g = 70.4 % F

47.9 g Zn= g Zn 1 mol Zn x Question 10: (Empirical formula) mol Zn 52.1 g Cl= g Cl 1 mol Cl x 1.47 mol Cl = 1 = 2 ZnCl 2 Assume 100g sample. 30 Zn Find the empirical formula of a compound, given that the compound is found to be: 47.9% zinc (Zn) and 52.1% chlorine (Cl) by mass. 17 Cl 35.45