Atoms, Molecules and Ions

Question

illustrate or show the solution:



1. Compound A contains 1.333 g of oxygen per gram of carbon, whereas compound B contains 2.666 g of oxygen per gram of carbon. What chemical law do these data illustrate?

2. Using the Law of multiple proportions, compare ethane (C2H6) and propane (C3H8). The weight of hydrogen which combines with 1g of C is 0.252 g in ethane and o.224 g in propane. Find its ratio in small numbers.

3. How many grams of carbon would be present in CO that contains 2.66 grams of Oxygen?

4. A student combines 154 grams of carbon tetrachloride and an unknown quantity of bromine in a sealed container to produce 243 grams of dibromodichlormethane and 71 grams of chlorine. How much bromine was used in the reaction, assuming the reactants are completely used up? Use the law of conservation of mass.

5. Using the law of definite proportions,

a. A 78.0-gram sample of an unknown compound contains 12.4 grams of hydrogen. Find the % by mass of hydrogen in the compound.

b. In compound XY, 3.5 grams of X reacts with 10.5 grams of Y. Find the % by mass of the X and Y elements respectively.

Answer 1

All five, with the working shown.

1. Which law do the carbon-oxygen data illustrate?

Compare the oxygen masses that combine with the same 1 g of carbon:

2.666 / 1.333 = 2.00, a ratio of 2:1 in small whole numbers

That is the Law of Multiple Proportions. When two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a simple whole-number ratio.

You can even identify them. In CO the ratio is 16.00/12.01 = 1.33 g O per g C, and in CO2 it is 32.00/12.01 = 2.67. So compound A is carbon monoxide and compound B is carbon dioxide, and the 2:1 ratio simply reflects one oxygen atom versus two.

2. Ethane versus propane

0.252 / 0.224 = 1.125 = 9/8

The ratio is 9:8.

Confirm it from the formulas. In C2H6 the hydrogen per carbon is 6/2 = 3 atoms; in C3H8 it is 8/3 atoms. Their ratio is 3 divided by 8/3 = 9/8. The experimental masses and the molecular formulas agree exactly, which is the point the law is making.

3. Carbon in CO containing 2.66 g of oxygen

CO is one carbon per oxygen, so the mass ratio is fixed at 12.01 to 16.00:

mass C = 2.66 g O x (12.01 g C / 16.00 g O) = 2.00 g of carbon

4. Bromine used, by conservation of mass

Total mass in must equal total mass out:

154 g + x = 243 g + 71 g = 314 g
x = 314 - 154

x = 160 g of bromine

Worth noticing how neatly this closes: 160 g is exactly 1 mol of Br2, 154 g is 1 mol of CCl4, 71 g is 1 mol of Cl2, and 243 g is 1 mol of CBr2Cl2. The reaction is CCl4 + Br2 giving CBr2Cl2 + Cl2, one mole of each. Whoever wrote the problem chose the numbers deliberately.

5a. Percent hydrogen by mass

(12.4 g / 78.0 g) x 100 = 15.9% hydrogen

5b. Percent composition of XY

Total mass = 3.5 + 10.5 = 14.0 g

%X = (3.5 / 14.0) x 100 = 25%
%Y = (10.5 / 14.0) x 100 = 75%

These must add to 100%, which is a free check on any percent-composition answer.

How the three laws differ, since the set is testing exactly that

Conservation of mass is about a single reaction: what goes in comes out. That is question 4.

Definite proportions is about a single compound: it always has the same composition by mass no matter how it was made or how much you have. That is question 5, and it is why a percentage answers the question at all.

Multiple proportions is about two or more different compounds of the same elements, and it is the only one of the three that requires atoms to exist. Whole-number ratios like 2:1 and 9:8 make no sense unless matter comes in indivisible units - which is precisely the argument Dalton used.