Mole Conversion Packet Short Version

Chemistry Name Key WS 1-Mole Conversion Packet-Short Version Date Block Calculate the formula mass (molar mass) for the following: Na 2 @ 22.989 = 45.978 S 1 @ 32.065 = 32.065 sodium sulfate Na₂SO₄ O 4 @ 16 = 64.00 142.043 g/mol calcium nitrate Ca(NO₃)₂ Ca 1 @ 40.078 = 40.078 N 2 @ 14.01 = 28.02 zinc acetate dihydrate Zn(C₂H₃O₂)₂ • 2H₂O O 6 @ 16 = 96 164.098 g/mol Calculate the percent composition of the following: carbon in carbonic acid aluminum in aluminum sulfate water in zinc phosphate tetrahydrate Convert the following: K: 125g 1 H₂SO₄ UK: mole H₂SO₄ 7) 125 g H₂SO₄ to moles [\frac{125g H₂SO₄}{1}] [\frac{1 mol H₂SO₄}{98.08 g H₂SO₄}] = 1.274 mol H₂SO₄   K: 100g KMnO₄ UK: moles KMnO₄ 100 g of KMnO₄ to moles [\frac{100g KMnO₄}{1}] [\frac{1 mol KMnO₄}{158.0 g KMnO₄}] = .63 mol KMnO₄ 35 g of CuSO₄ • 5H₂O to moles K: .5 mol H₂SO₄ UK: g H₂SO₄ 0.50 mole H₂SO₄ to grams [\frac{.5 mol H₂SO₄}{1}] [\frac{98.08 g H₂SO₄}{1 mol H₂SO₄}] = 49.04 g H₂SO₄ K: 1.7 mol KMnO₄ UK: g KMnO₄ 1.70 mole of KMnO₄ to grams [\frac{1.7 mol KMnO₄}{1}] [\frac{158.04 g KMnO₄}{1 mol KMnO₄}] = 268.67 g KMnO₄ 3.2 mole of CuSO₄ • 5H₂O to grams K: 1.5 x 10²⁰ Rep Part UK: mole 1.5 x 10²⁰ particles to moles [\frac{1.5 x 10²⁰ Rep Part}{1}] [\frac{1 mol}{6.02 x 10²³ Rep Part}] = .000249 mol Solve the following problems How many grams are there in (1.5 \times 10^{25}) molecules of CO₂? K: (1.5 \times 10^{25}) molecules CO₂ UK: g CO₂ [\frac{1.5 \times 10^{25} \text{ molecules CO₂}}{1}] [\frac{1 \text{ mol CO₂}}{6.02 \times 10^{23} \text{ molecules CO₂}}] [\frac{44.01 \text{ g CO₂}}{1 \text{ mol CO₂}}] = 1096.59 g CO₂ What volume would the CO₂ in problem 39 occupy? K: (1.5 \times 10^{25}) molecules CO₂ UK: L CO₂ [\frac{1.5 \times 10^{25} \text{ molecules CO₂}}{1}] [\frac{1 \text{ mol CO₂}}{6.02 \times 10^{23} \text{ molecules CO₂}}] [\frac{22.4 \text{ L CO₂}}{1 \text{ mol CO₂}}] = 558.14 L CO₂ A sample of NH₃ gas occupies 75.0 liters at standard conditions. How many molecules is this? K: 75 L NH₃ UK: molecules NH₃ [\frac{75 \text{ L NH₃}}{1}] [\frac{1 \text{ mol NH₃}}{22.4 \text{ L NH₃}}] [\frac{6.02 \times 10^{23} \text{ molecules NH₃}}{1 \text{ mol NH₃}}] = (2.015 \times 10^{24}) molecules NH₃ What is the mass of the sample of NH₃ in problem 41? K: 75 L NH₃ UK: g NH₃ [\frac{75 \text{ L NH₃}}{1}] [\frac{1 \text{ mol NH₃}}{22.4 \text{ L NH₃}}] [\frac{17.04 \text{ g NH₃}}{1 \text{ mol NH₃}}] = 57.05 g NH₃ How many atoms are in (1.3 \times 10^{22}) molecules of NO₂? K: (1.3 \times 10^{22}) molecules NO₂ UK: atoms NO₂ [\frac{1.3 \times 10^{22} \text{ molecules NO₂}}{1}] [\frac{3 \text{ atoms}}{1 \text{ molecule NO₂}}] = (3.9 \times 10^{22}) atoms in NO₂ A 5.0 g sample of O₂ is in a container at standard conditions. What is the volume of the container? K: 5 g O₂ UK: L O₂ [\frac{5 \text{ g O₂}}{1}] [\frac{1 \text{ mol O₂}}{32 \text{ g O₂}}] [\frac{22.4 \text{ L O₂}}{1 \text{ mol O₂}}] = 3.5 L O₂ How many molecules of O₂ are in the container in problem 44? How many atoms of oxygen? K: 5 g O₂ UK: molecules O₂ [\frac{5 \text{ g O₂}}{1}] [\frac{1 \text{ mol O₂}}{32 \text{ g O₂}}] [\frac{6.02 \times 10^{23} \text{ molecules O₂}}{1 \text{ mol O₂}}] = (9.406 \times 10^{22}) molecules O₂ 2