Chemistry 1010
Handout 1: Chemical substance and Measurements
1. A can of Campbell’s tomato soup indicates a suggested serving
of 10 fluid ounces or 274 mL.
(a) How many servings can be obtained from 1.00 litres of the
soup?
Servings =
1.00 L x 1000 mL L −1
274 mL
= 3.6 servings
(b) How many fluid ounces are there in 1.00 litre of the soup?
Fluid ounces in 1 L =
1.00 L x 1000 mL L −1 x 10 fluid ounces serving −1
274 mL serving −1
=
36 fluid ounces
(c) If the density of the soup is 0.9800 g mL–1, what is the
mass in grams of a serving?
Mass of one serving =
274 mL x 0.9800 g mL−1 = 269 g
2. Convert:
(a) 1.75 metres to centimetres.
1
(b) 30.15 inches to metres.
According to the data book that I have 1 in =
2.54 % 10 −2 m. The number of metres should be less
than the number of inches. So,
30.15 in x 2.54 % 10 −2 m in −1 = 0.76581 m. To the
correct number of significant digits the answer is = 0.766
m.
(c) 256 mL to L.
= 2.56 % 10 −1
L
Ans: 2.56 % 10 −1 L
256 mL
1000 mL L −1
(d) 3.61 L to mL
1 L = 1000 mL so there must be more mL that L!
Ans: 3.61 % 10 3 mL
(e) 683.4 cm3 to dm3. 1 dm = 10 cm so 1 dm3 = (10 cm)3 =
1000 cm3. So,
683.4 cm3
683.4 cm3 = 1000 cm3 dm−3
:Ans
= 6.834 % 10 −1 dm3
3. Recording tape in a cassette player travels at 1 7/8 inches per
second. What is the length of the tape in a cassette if it is
measured in metres and plays for 45 min per side (C-90)?
Length of tape in inches =
15
8
in s −1 x 60 s min -1 x 45min = 5062.5 in
Length of tape in metres = 5062.5 in x 2.54 cm in-1 x
1
1000 cm m −1
Volume in litres = 250 imperial_gallon x 4.545 L
imperial_gallon-1 = 1.13625 x 103 L
(b) What is the internal length of the sides of the cube in cm?
Volume of the cube = 1.13625 x 103 L x 1000 cm3 L-1 =
1.13625 x 106 cm3
The volume of a cube = (length of side)3
Length of side = 3
1.13625 x 10 6 cm3 = 104.3497 cm
(c) What is the total area of the steel sheet needed to make
the cube?
Each face has an area = (side)2.
There are 6 faces so the total area = 6 x (104.3497 cm)2 =
6.53 x 104 cm2
5. Evaluate the following expressions giving your answer to the
There are 1 centimetre is 100 metre so there are 100 cm
in one metre and the number of centimetres will be greater
than the number of metres.
1.75 m x 1000 cm m−1= 1750 cm but this leaves the
number of significant digits in doubt. It is better to quote
the answer in scientific notation so the answer becomes (3
significant digits)
Ans : 1.75 x 10 3 cm
There are 1000 mL in 1 litre so 256 mL =
(a) What is the required volume in litres?
= 12.86 m
Ans: 13 m
4. A company wishes to make fuel tanks in the shape of a regular
cube (all sides have same length) to hold 250 imperial gallons
when completely filled. (Note: 1 imperial gallon = 4.545 L).
correct number of significant digits.
(a) 2.673 - 1.71 x 10–3 + 3.3216 x 102
334.83
(b) 232.81 + 2.221 + 0.0335 + 0.000044
235.06
(c)
12.226 x (1.312 + 1.35 x 10 2 )
(273.412 − 101.3) x 1.732
5.59
6. In a kitchen you will usually find a few pure substances.
Calculate the formula weight and the molar mass of each of the
following:
(a) 'Common salt': sodium chloride, NaCl
58.4425 u; 58.44245 g.mol–1
(b) 'Baking soda': sodium hydrogen carbonate, NaHCO3
84.007 u; 84.007 g.mol–1
(c) 'Lye': sodium hydroxide, NaOH
39.9971; 39.9971 g.mol–1
(d) 'Sugar': sucrose, C12H22O11
39.99707; 39.99707 g.mol–1
(e) 'Bleach': sodium hypochlorite, NaOCl
74.4418; 74.4418 g.mol–1
(f) 'Washing soda': sodium carbonate decahydrate,
Na2CO3·10 H2O
286.141; 286.141 g.mol–1
(g) 'Vinegar': ethanoic acid (as an aqueous solution), C2H4O2.
60.052; 60.052 g.mol–1
7. Balance the following equations:
(a) 2 C(s) + O2(g) → CO(g)
(b) 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
(c) CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
(d) WO2(s) + 2 H2(g) → W(s) + 2 H2O(g)
8. How many grams of the named substance is contained in the
stated number of moles?
(a) 0.50 mol of Ag(s)
Ans: 54 g
(b) 5.032 mol Ca(NO3)2
(c) 1.01 x 10–3 mol CuSO4·5 H2O
Ans: 825.7 g
Ans: 0.252 g
9. How many moles of the named substance is contained in the
stated mass?
(a) 50.0 g of S(s)
Ans: 1.56 mol
(b) 0.05000 g of NaCl
(c) 22.26 g of (NH4)2SO4
Ans: 0.0008555 mol
Ans: 0.1951 mol