Every card in Quantitative chemistry, Year 10: formula mass, uncertainty and moles

Relative formula mass (Mr)

HintIts symbol is a capital M with a small r.

WhyFor water, H2O, it is 1 + 1 + 16 = 18. It has no unit because, like relative atomic mass, it is a comparison of masses and not a mass in grams.

58

HintThe 2 outside the bracket applies to everything inside it.

WhyThe formula contains one Mg, two O and two H: 24 + (2 × 16) + (2 × 1) = 58. Work out the bracket first, 16 + 1 = 17, then double it and add the magnesium.

28; 35

HintFind the part that is nitrogen, then express it as a fraction of the whole.

WhyThere are two N atoms in the formula, one in NH4 and one in NO3, so nitrogen contributes 2 × 14. Dividing that by the relative formula mass and multiplying by 100 gives the percentage: 28 ÷ 80 × 100.

relative formula mass

HintYou are comparing a part with the whole, so divide by the total for one unit of the compound.

WhyThe top of the fraction is the Ar of the element multiplied by the number of its atoms in the formula. The bottom is the Mr of the whole compound. AQA expects you to be able to do this calculation when given the relative masses.

44, because the totals on each side of a balanced equation are equal

HintNo atoms are made or lost, so whatever goes in must come out.

WhyIn a balanced equation the relative formula masses of the reactants, in the amounts shown, add up to the same total as those of the products: 100 = 56 + 44. You can check it from the atoms: 12 + (2 × 16) = 44.

21.1; 0.6; 0.3

HintAdd and share out for the first; subtract smallest from largest for the second; then split that in two.

WhyMean = (21.0 + 21.4 + 20.8 + 21.2) ÷ 4. Range = 21.4 − 20.8. The readings lie within half the range either side of the mean, so the result can be written as the mean ± half the range.

uncertainty

HintIt is the doubt about how close a reading is to the true value.

WhyNo instrument and no person reading it is perfect, so a measured value is never known exactly. Scientists estimate how big the doubt is and report it with the result.

Group A's, because its readings have the smaller range

HintCompare how widely each set is scattered around the middle.

WhyGroup A's readings span 0.4 cm³, so its mean is 25.0 ± 0.2 cm³. Group B's span 4.0 cm³, so its mean is 25.0 ± 2.0 cm³. The more closely the repeats agree, the more confident you can be in the mean.

The true value is likely to lie between 24.8 and 25.2 cm³

HintDo the sum once with a plus and once with a minus.

WhyThe figure after the ± sign is the uncertainty. It gives the interval around the measured value within which the true value is expected to lie: from 25.0 − 0.2 up to 25.0 + 0.2.

One reading cannot show how much the results vary

HintA spread needs at least two points.

WhyThe uncertainty is estimated from how far the repeat readings are scattered about their mean. With a single reading there is no scatter to look at, so there is no way to judge it.

The Avogadro constant

HintIt is named after an Italian scientist, and its value is about 6 followed by 23 zeros.

WhyIts value is 6.02 × 10²³ per mole. One mole of any substance contains this number of the particles named.

2

HintFirst find the mass of one mole from the formula.

WhyThe relative formula mass of CO2 is 12 + (2 × 16) = 44, so one mole has a mass of 44 g. Number of moles = mass ÷ Mr = 88 ÷ 44.

40; 10

HintAdd up the three atoms, then remember that one mole has that many grams.

WhyMr = 23 + 16 + 1. The mass of one mole in grams is numerically equal to the relative formula mass, so mass = moles × Mr = 0.25 × 40.

3.01 × 10²³

HintHalf a mole holds half of what a whole mole does.

WhyNumber of particles = moles × the Avogadro constant = 0.5 × 6.02 × 10²³. Only the 6.02 is halved; the power of ten stays the same.

Neither: they contain the same number

HintA mole is a counting unit, like a dozen.

WhyOne mole of anything contains the Avogadro constant of the particles named: here atoms in one case and molecules in the other. The mole of carbon dioxide has a greater mass (44 g against 12 g) because each of its particles is heavier.