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Chemistry 1010 Sections 1, 2, 3 and10 Loader (Fall 2004)
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Chemistry 1010
Handout 2: Moles and Stoichiometry - 1
1.
A 10.0 g sample of tellurium oxide is found, on analysis, to contain 8.0 g of tellurium. What is the empirical formula of the oxide?
The amount of oxygen present in the sample = 10.0 - 8.0 g = 2.0 g
Te
O
Mass in
sample
8.0 g
2.0 g
Moles in
sample
8.0 g
127.60 g mol −1 = 0.0627 mol
2.0 g
15.9994 g mol −1 = 0.125 mol
Ratio
0.0627 mol
0.0627 mol = 1.0
0.125 mol
0.0627 mol = 2.0
So the empirical formula is Te02
2.
An oxide of chlorine contains 18.4% of oxygen and 81.6% of chlorine. What is the empirical formula of the oxide?
ClO2
3.
Calculate the empirical formula for the compound which has the following percentage composition by weight: C, 61.0%; N, 23.6%;
H, 15.4%
C3NH9
4.
The chief ore of mercury is cinnabar, HgS. Pliny, in 77 A.D. reported that the 10,000 lb of this ore was imported to Rome annually
from Spain. Assuming the ore to be pure mercury(II) sulfide, how many lb of mercury could be recovered from 10,000 lb of the ore?
I mole HgS → gives 1 mole of Hg. HgS molar mass = 232.656 g mol-1 and Hg molar mass 200.59 g mol-1
200.59 g mol −1
Maximun mass of mercury recovered = 232.65 g mol −1 x 10000 lb= 8621.7 lb
Ans.
6
5.
The equation for the reaction of concentrated sulfuric acid with sodium chloride is
H2SO4(l) + NaCl(s) → NaHSO4 + HCl(g).
What mass of hydrogen chloride would be obtained when 234 g of sodium chloride reacts with excess sulfuric acid?
First check that the equation is balanced - it is. I mole of NaCl gives 1 mole of HCl.
234 g
Moles of NaCl available = 58.4425 g mol −1 =4.004 mol
Maximum mass of HCl = 4.004 mol x 36.4606 g mol-1 = 146.0 g
Ans
6.
Tetraphosphorus hexaoxide, reacts with iodine according to the equation:
5 P4O6 + 8 I2 → 4 P2I4 + 3 P4O10
(a) What mass of P4O6 would be needed to react exactly with 1.00 g of I2?
Check for equation balance - OK. Focus on the I2 :
3
5
8 P4O6 + 1 I2 → ½ P2I4 + 8 P4O10
1.00 g
5
Moles of I2 = 126.9045 g mol −1 =0.007880 mol s o moles of P406 required = 8 x 0.007880 mol = 0.0049250 mol
So, the mass of P406 required = 0.0049250 mol x 219.8916 g mol-1 = 1.08 g Ans
(b) How many mole of P4O6 would be consumed in the production of 5.4 g of P2I4?
3
5
4 P4O6 + 2 I2 → 1 P2I4 + 4 P4O10
Ans: 2.6 g
7.
1.000 g of pure PbCl2 is converted to 1.031 g of AgCl. Find the atomic mass of lead. (Show all workings).
Start by writing the equation and balacing it.
Ans: 207.1 u
8.
An unknown element forms a sulfate formula XS04. When 0.7550 g of the sulfate is dissolved in water and a solution of barium
nitrate is added, a precipitate of barium sulfate is formed. When collected and dried the precipitate of barium sulfate weighs 1.1601
g. Calculate the atomic weight of X.
The reaction is XS04(aq) + Ba(NO3)2(aq) → BaS04(s) + X(NO3)2(aq). I mole of the sulfate give 1 mole of BaS04(s)
1.1601 g
Moles of BaS04(s) formed = 233.39 mol L −1 = 0.0049706 mol. So the number of moles of XS04 also = 0.0049706 mol
0.7550 g
Molar mass of XS04 = 0.0049706 mol = 151.893 g mol-1. Now the SO4 part has molar mass = 95.95173 g mol-1 so
molar mass of X = 151.893 g mol-1 - 95.95173 g mol-1 = 40.01 g mol-1 Ans.
Chemistry 1010 Sections 1, 2, 3 and10 Loader (Fall 2004)
9.
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Balance the following equations by inspection:
(a) 2 CO(g) + O2(g) → 2 CO2(g)
(b) CS2(l) + 3 Cl2(g) → CCl4(l) + S2Cl2(l)
(c) Al2O3(s) + 6 HCl(aq) → 2 AlCl3(aq) + 3 H2O(l)
(d) CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2(g)
10. Write balanced chemical equations for the following reactions:
(a) aluminum metal + oxygen gas → solid aluminum oxide
4 Al(s) + 3 O2(g) → 2 Al2O3(s)
(b) aqueous silver sulfate + aqueous barium iodide → solid silver iodide + solid barium sulfate
Ag2SO4(aq) + BaI2(aq) → 2 AgI(s) + BaSO4(s)
(c) methane gas + oxygen gas → carbon dioxide gas + water(l)
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
(d) aluminum metal + aqueous hydrochloric acid → aqueous aluminum chloride + hydrogen gas.
2 Al(s) + 6 HCl(aq) → 2 AlCl3 + 3 H2(g)
11. Analysis shows that a 0.500 g sample of a compound of phosphorus and chlorine contains 0.07437 g of phosphorus and 0.4260 g
of chlorine. Calculate the empirical formula for the compound?
PCl5
12. Elemental analysis of glucose (a sugar) gives the following results:
C, 40.02; H, 6.66; 0, 53.31% (by weight)
Calculate (a) the empirical formula and (b) the molecular formula if the molar mass is 180.
(a) CH2O (b) C6H12O6
13. Once upon a time I prepared a compound with the expected molecular formula, C6H5NO2. Calculate the percentage composition by
weight for this compound.
C 58.54%, H 4.09%, N 11.38%, O 25.99%
14. A sample 0.7505 g of a solid containing only C, H and 0 was burned completely to give 1.8940 g of CO2, and 0.3320 g of H2O in the
reaction:
oxygen + 0.7505 g of solid → 1.8940 g CO2 + 0.3320 g H2O
Calculate the empirical formula of the compound.
Some working - not the worked problem!
12.011
x 1.8940 g = 0.516904 g
Mass of C present in sample = 12.011 + (2 x 15.9994)
2 x (1.0079)
Mass of C present in sample = 15.9994 + (2 x 1.0079)
x 0.3320 g = 0.0371489 g
Mass of O = 0.7505 - (0.516904 g + 0.0371489) g = 0.196447 g
C7H6O2
15. Balance the following simple equations:
(a) N2(g) + 3 H2(g) → 2 NH3(g)
(b) 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
(c) Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
16. A solution contains 11.26 g of magnesium chloride, MgCl2, per 1.00 L of solution.
(a) What is the molarity of the MgCl2 solution?
0.118 mol L-1
(b) How many moles of MgCl2 are contained in 50.00 mL of the solution?
5.90 x 10-3 mol
17. A doctor's prescription calls for a solution of potassium chloride, containing 1.49 g of potassium chloride per 15 cm3 of solution (his
ampoules hold 15 cm3). What is (a) the concentration of this solution in mol cm-3 and (b) what amount of potassium chloride must
be weighed to make 1.000 dm3 of solution?
(a) 1.33 x 10-3 mol cm-3 (b) 99.3 g
18. In the reaction
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaCl(aq)
What is the (a) number of mol and (b) the mass in g of silver chloride that can be made from 25.00 cm3 of 0.1272 mol L-1 NaCl?
(a) 3.180 x 10-3 mol (b) 0.4558 g
Chemistry 1010 Sections 1, 2, 3 and10 Loader (Fall 2004)
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19. The chloride, XCl2, of an unknown element X, reacts with silver nitrate, AgNO3 as follows:
XCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + X(NO3)2(aq)
25.00 mL of a solution containing 0.9248 g of XCl2 gave 1.2730 g of AgCl(s).
Calculate:
(a) the number of mol of AgCl(s) formed,
0.008882 mol
(b) the number of mol of XCl2 present in the 25.00 mL of solution,
0.004441 mol
(c) the molar mass of the XCl2,
208.2 g mol-1
(d) the atomic weight of X, and
137.3
(e) the concentration (in mol L-1) of the XCl2 solution.
0.1776 mol L-1
20. A solution of sulfuric acid is described as 97% sulfuric acid by weight. What (a) is the concentration in mol L-1 and (b) what volume
is required to make 500 mL of 6 mol L-1 solution? The density of 97% H2SO4 is 1.84 g mL-1
(a) 18 mol L-1 (b) 0.17 L