Exercise practice: Some Basic Concepts of Chemistry

4.1 How the exercises are grouped

The chapter has exercises 1.1 to 1.36. They are not thirty-six unrelated problems: almost all of them test one of five skills. The groups below name the exercises and then work a representative question in full. Where a group has many similar parts, the method shown covers the rest.

4.2 Group A: formula mass and composition

Exercise 1.1 asks for the molar masses of water, carbon dioxide and methane. Add the atomic masses in the formula and attach the unit g mol^-1. Water gives 2(1.008) + 16.00 = 18.02 g mol^-1, carbon dioxide gives 12.011 + 2(16.00) = 44.01 g mol^-1, and methane gives 12.011 + 4(1.008) = 16.04 g mol^-1.

Exercise 1.2 asks for the mass percentage of the elements in sodium sulphate. The molar mass is 2(23.0) + 32.06 + 4(16.00) = 142.06 g mol^-1. Sodium contributes 46.0 g, so its mass percentage is 46.0 divided by 142.06, which is 32.4 percent. Sulphur contributes 32.06 g, which is 22.6 percent, and oxygen contributes 64.0 g, which is 45.0 percent. The three percentages add to 100.

Exercises 1.3 and 1.8 use the same data: an oxide of iron contains 69.9 percent iron and 30.1 percent oxygen by mass. Take 100 g of the compound. Iron gives 69.9 divided by 56.0, which is 1.25 mol, and oxygen gives 30.1 divided by 16.0, which is 1.88 mol. Dividing both by the smaller value gives 1 for iron and 1.5 for oxygen. Multiply by two to reach whole numbers: 2 iron and 3 oxygen. The empirical formula is Fe2O3, and its formula mass is 2(56.0) + 3(16.0) = 159.7 u, which is the same as the molar mass quoted for the oxide, so the molecular formula is also Fe2O3.

Exercise 1.9 works the average atomic mass of chlorine from the two isotopes. Multiply each isotopic mass by its fractional abundance and add: 34.9689 x 0.7577 plus 36.9659 x 0.2423 gives 35.45 u. Exercise 1.32 repeats the method for the three argon isotopes and gives about 39.95 u. Exercise 1.34 uses combustion data to get the empirical formula CH, then the measured mass of 10.0 L at STP gives a molar mass near 26 g mol^-1, so the molecular formula is C2H2.

4.3 Group B: units and significant figures

Exercise 1.13 converts a pressure given as a mass per unit area. The air mass is 1034 g per square centimetre, which is 10.34 kg m^-2 after converting grams to kilograms and square centimetres to square metres. Multiplying by the acceleration due to gravity gives about 1.01 x 10^5 Pa, close to the standard atmospheric pressure.

Exercise 1.22 multiplies the speed of light by the time: 3.0 x 10^8 m s^-1 multiplied by 2.00 x 10^-9 s gives 0.60 m. Exercise 1.27 converts 28.7 pm and 25365 mg into base units, giving 2.87 x 10^-11 m and 2.5365 x 10^-2 kg. Exercise 1.30 finds the mass of one carbon-12 atom by dividing 12 g by the Avogadro constant, which gives 1.99 x 10^-23 g.

For significant figures, keep the rule and the count together. In 0.02856 x 298.15 x 0.112 divided by 0.5785, the factor 0.112 has three significant figures and the others have four or more, so the answer keeps three. In 5 x 5.364 the number 5 has a single significant figure, so the answer is reported as 3 x 10^1. In 0.0125 + 0.7864 + 0.0215 the limiting piece is the number of decimal places, four, so the sum keeps four decimal places.

4.4 Group C: moles, particles and concentration

Exercise 1.10 asks for the atoms in three moles of ethane. Each molecule C2H6 has six carbon atoms and eighteen hydrogen atoms in all, so three moles contain 6 mol of carbon atoms, 18 mol of hydrogen atoms and 3 x 6.022 x 10^23 = 1.807 x 10^24 molecules.

Exercise 1.11 dissolves 20 g of sugar in enough water to make 2 L. The molar mass of C12H22O11 is 342.3 g mol^-1, so 20 g is 0.0584 mol and the molarity is 0.0584 divided by 2, which is 0.0292 M. Exercise 1.12 reverses the calculation for methanol: 2.5 L of 0.25 M solution needs 0.625 mol, which has a mass of 20.0 g, and at a density of 0.793 kg L^-1 that is a volume of 25.2 mL.

Exercise 1.28 asks which 1 g sample holds the most atoms. Divide 1 g by the molar mass and count atoms per formula: 1 g of lithium is the largest number of moles per gram among the options, so lithium holds the most atoms. Exercise 1.29 converts a mole fraction into a molarity. A mole fraction of 0.040 for ethanol means 0.040 mol of ethanol with 0.960 mol of water; the water mass is 17.3 g, so the solution volume is about 17.3 mL and the molarity is about 2.3 M.

4.5 Group D: stoichiometry and limiting reagent

Exercise 1.24 burns dinitrogen with dihydrogen to make ammonia. Convert both masses to moles, divide each by its coefficient in the balanced equation, and the smaller quotient decides the limiting reagent. With 2.00 x 10^3 g of nitrogen and 1.00 x 10^3 g of hydrogen, hydrogen is in short supply, so the ammonia is calculated from hydrogen and some hydrogen remains unreacted.

Exercise 1.35 reacts calcium carbonate with hydrochloric acid. The acid sample is 25 mL of 0.75 M solution, which is 0.01875 mol of HCl. The equation needs two moles of acid for each mole of carbonate, so the carbonate needed is 0.009375 mol, and at 100.1 g mol^-1 the mass is 0.94 g. Exercise 1.36 works the same way with the manganese dioxide reaction and gives the mass of hydrogen chloride that reacts with 5.0 g of manganese dioxide.