10. Even though lead is to lead compounds were used in ancient times as white pi

Question

10. Even though lead is to lead compounds were used in ancient times as white pigments in cosmetics. What is the percentage of lead by mass in lead(IV) carbonate, Pb(CO3)? a. 81.4% d. 77.5% b, 63.3% e. 32.7% c. 20.7% 11. A hydrocarbon molecule contains carbon and hydrogen atoms in equal numbers. Its molar mass is 130.18 glmol. What is the molecular formula for the hydrocarbon? d. C12H12 a. CH e. C10H10 b. C8Hs c. C1H7 12. Elemental analysis of the soot produced by a candle flame shows that it is 14.3% H and 85.7% C by mass. What is the empirical formula of this hydrocarbon? d, CH a. CH2 b. C2H e. CH3 c. C2H3

Explanation / Answer

10. % Lead:

The molar mass of the Lead (IV) carbonate [Pb(CO3)2] is 327,2178 g/mol.

The atomic weight of the lead (Pb) is 207,2 g/mol.

% Composition = (atomic weight of element / molar mass of Compound) x 100

%Lead = (207,2 / 327,2178)x100= 63,32%

11. Molecular formula: mass of the hydrocarbon is 130,18 g/mol

It says the hydrocarbon molecule has equal numbers of C and H, so the general formula is CnHn, where “n” will be the number of atoms of C and H. Also we have to know the atomic weith of C, which is 12g/mol and the atomic weith of H, which is 1g/mol.

So, the molar weith of the hydrocarbon (M) will be: M= 12.n + 1.n

130,18 = 12.n + 1.n

Clearing the n, the value is: 10

So, as the value of n is 10 and the general formula of our hydrocarbon molecule is CnHn, the actual molecular formula es C10H10

12. Empirical formula: 14,3% H and 85,7% C by mass

We need to assume we have 100 g of hydrocarbon, and we need to change grams to moles:

With the next formula we have the grams of the atoms:
% Composition = (Mass of element / Mass of Compound) x 100

Grams of H = 14,3g

Grams of C = 85,7g

With the atomic weights we already know, we calculate the moles of H and C:

Moles of H = 1 mol/1g x 14,3g = 14,3 = 14 moles of H

Moles of C = 1 mol/12g x 85,7g = 7,1417 = 7 moles of C

To find the empirical formula of the compound is needed to divide each value obtained by the lower value between them, which is 7. So we obtain: CH2

ANSWERS:

10. Option B

11. Option E

12. Option A

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