A sample of compound containing boron and hydrogen contains 6.444 g of B and 1.803 g of H. The molar mass of the compound is about 30 g. What is its empirical and molecular formula?...thank you for sharing your knowledge...

Answer 1

1. convert the masses to the mole using the molar mass;

n(B) = m/Mr = 6.444g/10gmol-1 = 0.6444mol

n(H) = m/Mr = 1.803g/1gmol-1 = 1.803mol

divide the moles by the smallest mole;

0.6444/0.6444 = 1 B

1.803/0.6444 = 2.79 H

Use the whole number as we are to find the empirical formula so 2.79 becomes 3.

so the compound contains 1mol of B and 3 mol of H
i.e. BH3

z = Mr/x where x is the molar mass of the empirical formula and z is the mole in whole number;
B = 10g/mol
H = 1g/mol x 3 = 3g/mol

Mr(BH3) = 13g/mol

z = 2.307 = 2

Molecular formula = z(BH3) = 2(BH3) = B2H6

***Note that i used the whole number molar masses and i end up with decimals which i rounded them off to the whole number. Try use the molar mass up to 2 sig figures from the periodic table which will help you get the whole number figures***

hope that helps

Answer 2

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