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Question

1-What is the subscript for the hydrogen in a molecular formula of a compound containing only carbon and hydrogen which upon combustion produced 1.55 g CO₂ and 0.762 g H₂O? The molar mass of a molecular compound is 216.4 g/mol. 

2-What is the molar concentration (molarity) of (4.040x10^2) mL of solution made from (2.58x10^0) grams of oxalic acid dihydrate? 

Answer 1

1.) When a compound composed of only C and H is combusted, all of the carbon (C) ends up as CO2 and the hydrogen (H) ends up in H2O. So, we find the moles of CO2 and H2O, and from that we find moles of C and H in the compound.

moles C: 1.55 g CO₂ x 1 mol CO₂ / 44.01 g x 1 mol C / mol CO₂ = 0.0352 mols C

moles H: 0.762 g H₂O x 1 mol H₂O / 18.02 g x 2 mol H / mol H₂O = 0.0846 mols H

Now, to get whole numbers, we divide both by 0.0352:

mols C = 0.0352 / 0.0352 = 1 mol C

mols H = 0.0846 / 0.0352 = 2.4 mols H

Still not whole numbers, so we multiply by 5 to get 5 mols C and 12 mols H

Empirical formula would then be C₅H₁₂ with a molar mass of 5x12 + 12x1 = 60 + 12 = 72

A molecular compound with a molar mass of 216 would then be 216 / 72 = 3 empirical formulae

Molecular formula would then be C₁₅H₃₆

Subscript for hydrogen = 36

2.) 2.58x10⁰ g oxalic acid = 2.58 g oxalic acid

molar mass oxalic acid (H₂C₂O₄) = 90.03 g/mol

moles oxalic acid = 2.58 g x 1 mol / 90.03 g = 0.02866 mols

4.040x10² ml = 404.0 mls = 0.4040 L

molarity = moles oxalic acid / L solution

molarity = 0.02866 mols / 0.4040 L = 0.07093 M = 0.0709 M (3 sig. figs.)