1. (18 marks)
(a) (3 marks) Write the formula of the compound formed between:
sodium and sulfur Na2S
aluminum and fluorine AlF3
calcium and phosphorus Ca3P2
(b) (3 marks) Round the following calculations to the proper number of significant figures:
9527.3 + 575.4847 + 2.356 = 10105.1407 10105.1
(1.535 x 36.0045) / 165.72 = 0.33349570 0.3335
(54.785 - 49.186) / 186.2073 = 0.03055197 0.03055 (54.785 - 49.186 = 5.599; 4 s.f.)
(c) (3 marks) Write formulas for the following compounds:
cobalt(II) carbonate CoCO3
aluminum nitrate Al(NO3)3
ammonium sulfate (NH4)2SO4
(d) (3 marks) Name the following compounds:
CH4 methane
P4O10 tetraphosphorus pentaoxide
FeCl3 iron(III) chloride
(e) (6 marks) State Dalton's Law of Multiple Proportions. Show how it applies to the compounds IF (86.98% I) and IF5 (57.19% I).
When 2 elements form two different compounds, the mass ratio of the elements in one compound is a small whole-number ratio of the mass ratio of the elements in the other compound.
IF: 86.98 g I / 13.02 g F = 6.680
IF5: 57.19 g I / 42.81 g F = 1.336IF mass ratio / IF5 mass ratio = 6.680 / 1.336 = 5.001 = 5.000 within experimental error (3rd decimal place).
Thus IF has 5 times as much I for a given mass of F as IF5.
(Note: The 5:1 ratio of the I:F mass ratios in these two compounds is what led Dalton to the idea that one of these compounds contained five times as many atoms of F per atom of I as the other compound.)
2. (10 marks) Fill in the blanks in the following questions:
(a) Consider the ion
The number of protons in this ion is 33. The number of neutrons is 42. The number of electrons is 36 . The atomic number is 33. The mass number is 75. The element is As (arsenic).
(b) A mole of boron atoms contains 6.022 x 1023 atoms. A mole of B12 molecules contains 6.022 x 1023 molecules. A mole of B12 molecules contains 72.264 x 1023 atoms. A mole of B12 molecules has a mass of 129.732 grams.
3. (10 marks) A compound contains 68.55 % C, 8.63 % H, and 22.83 % O. What is its empirical formula?
Get moles of each element, then determine the mole ratio. Assume 100 g of the compound, so that you have 68.55 g C, 8.63 g H, 22.83 g O. 1 extra digit shown in calculations below.
68.55 g C x (1 mol C / 12.011 g C) = 5.7073 mol C
8.63 g H x (1 mol H / 1.0079 g) = 8.562 mol H
22.83 g O x (1 mol O / 15.999 g O) = 1.4270 mol O
Moles of N is the smallest number; get other ratios relative to O:
5.7073 mol C / 1.4270 mol O = 4.000 mol C / mol O
8.562 mol H / 1.4270 mol O = 6.000 mol H / mol O
Therefore the empirical formula is C4H6O.
(b) If the experimental molar mass is 140.2 g/mol, what is the molecular formula?
C4H6O has a formula weight of 4*12.011 + 6*1.0079 + 15.999 = 70.09 g/mol. This is exactly half of the experimental molar mass (140.2 / 70.09 = 2.000), so the molecular formula is C8H12O2.
4. (10 marks) A sample was analyzed for alum, KAl(SO4)2. The sample was dissolved in water, excess BaCl2 was added to precipitate BaSO4, and the BaSO4 was filtered and weighed. If 0.4720 g of the sample yielded 0.6416 g of BaSO4, what is the % KAl(SO4)2 in the sample? The balanced equation is
First determine the mass of KAl(SO4)2 that can be produced from 0.6416 g BaSO4:
= 0.35489 g KAl(SO4)2 (1 extra digit shown; all digits carried in calculator)
% Al = 0.35489 g KAl(SO4)2 / 0.4720 g x 100 = 75.19%
5. (12 marks) Arsenic reacts with oxygen to produce diarsenic trioxide according to the following equation:
(a) How many moles of oxygen are required to react with 125 grams of arsenic?
= 1.25 mol O2
(b) What mass of diarsenic trioxide can be produced?
= 165 g
(c) If 158 g of As2O3 are actually obtained, what is the percent yield of As2O3 ?
158 g obtained/ 165 g possible x 100
= 95.8%
6. (18 marks)
(a) (10 marks) Balance the following equations, using the lowest possible whole-number coefficients:
2 SO2 + O2 ---> 2 SO3
2 Ag + 2 HNO3 ---> 2 AgNO3 + H2
I2 + 7 F2 ---> 2 IF7
2 AuCl3 + 3 Na2S ---> Au2S3 + 6 NaCl
2 C3H6 + 9 O2 ---> 6 CO2 + 6 H2O
(Note: coefficients of "1" are not normally written. If there is no coefficient, it is understood to be "1").
(b) (3 marks) Gold metal has a density of 19.3 g/cm3. What is its density in lb/ft3? (Conversions: 1 lb = 454 g; 1 ft = 30.48 cm)
19.3 g / cm3 x (1 lb / 454 g) x (30.48 cm / ft)3 = 1.20 x 103 lb/ft3
NOTE: (30.48 cm / ft)3 = (30.48 cm / ft) x (30.48 cm / ft) x (30.48 cm / ft)
(c) (5 marks) The mass of 0.2750 mol of a compound with the formula MO3 is 63.76 g.
(i) What is the molar mass of MO3?
63.76 g / 0.2750 mol = 231.9 g/mol
(ii) What is the atomic weight of M?
231.9 g/mol MO3 - 3 mol O x (15.9994 g O /mol O) = 183.9 g/mol M
(iii) What is M? (give the symbol).
The element with atomic weight closest to 183.9 is W (tungsten)
7. (10 marks) Answer the following questions using the figures shown here.
NOTE: Each question may have more than one answer; each figure may be used more than once. Write "NONE" if no drawing fits the description.
(i) Which drawing(s) represent submicroscopic particles in a sample of a heterogeneous mixture? (c)
(ii) Which drawing(s) represent submicroscopic particles in a sample of an ionic solid? (d)
(iii) Which drawing(s) represent submicroscopic particles in a sample of a compound? (d), (f)
(iv) Which drawing(s) represent submicroscopic particles in a gaseous mixture? (b)
8. (12 marks) You are a demonstrator in the CHEM 1P80 class and you need to prepare 7.000 L of a 20.00% (by mass) solution of sodium chloride. Use the graph below to help you answer the following questions:
(a) What is the density of a solution that is 20.00 wt. % NaCl?
1.147 +/- 0.003 g / cm3
(Full units are g solution / cm3 solution)
(b) What mass (in grams) of sodium chloride would you take to make up the solution?
mass solution = (7.00 L solution ) x (1000 mL/L) x 1.145 g solution /mL solution
= 8015 g solution
8015 g solution x (20.00 g NaCl /100 g solution)
= 1603 g NaCl
(c) What mass (in grams) of water would you take to make up the solution?
mass water = mass solution - mass NaCl
= 8015 g solution - 1603 g NaCl
= 6412 g water
This page is
http://chemiris.labs.brocku.ca/~chemweb/courses/chem180/Midterm_1B_2000.html
Created October 17, 2000 by M. F. Richardson
© Brock University, 2000
