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Worked solutions to student book questions

Chapter 2 Analysis by mass
Q1.
a b Why was the soup sample in Worked Example 2.1 heated to 110°C?
Why was it necessary to weigh the sample four times?

A1. a b

The soup was heated above 100°C to evaporate water from the sample.
By repeatedly heating the sample until the mass remained unchanged, the analyst could be sure that all the water had been removed.

Q2.
Some laboratories use microwave ovens in place of conventional ovens to dry samples. What advantage could this have?
A2.
Microwave ovens dry samples more rapidly than conventional ovens.
Q3.
Soy sauce weighing 74.6 g was heated in an oven to constant mass. The final mass was 14.2 g. What percentage of water did the sauce contain?
A3.
Mass water = 74.6 – 14.2 = 60.4 g
100 60.4 100 mass of water
=
×
= 80.97 = 81.0%
×
%Water = mass of sample
1
74.6
1
(three significant figures)
Q4.

Three brands of dog food were heated and dried to constant mass. The data recorded are shown in Table 2.2.
Table 2.2 Determination of water content of dog food
Dog food tested
Phydeaux Deluxe
K9 Budget
Fresh meat – buffalo mince a b

Mass of dog food sample (g)
19.8
7.4
15.0

Final mass (g)
3.9
1.9
3.8

Which brand contained the highest percentage of water?
Do you consider that water content is a good guide to the relative value of different dog foods? What other factors might be important?

Heinemann Chemistry 2 4th edition Enhanced
Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd)

1

Worked solutions to student book questions

Chapter 2 Analysis by mass
A4.
a

b

19.8 − 3.9 100
= 80%
×
19.8
1
7.4 − 1.9 100
K9 Budget
% water =
= 74%
×
7.4
1
15.0 − 3.8 100
Fresh meat – buffalo mince % water =
= 75%
×
15.0
1
Water would not be a good guide to the nutritional value of the dog food. The amount of protein, carbohydrates, fats, vitamins and minerals need to be

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