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AlladinOne [14]
3 years ago
6

A compound has a percent composition of 54.5% carbon, 9.3% hydrogen and 36.2 % oxygen.If its molar mass is 88 g/mol, what is its

molecular formula?
Complete the following:

Chemistry
1 answer:
drek231 [11]3 years ago
4 0

Answer:

C₄H₈O₂.

Explanation:

  • Firstly, we can calculate the no. of moles (n) of each component using the relation:

<em>n = mass/atomic mass,</em>

mol C = mass/(atomic mass) = (54.5 g)/(12.0 g/mol) = 4.54 mol.

mol H = mass/(atomic mass) = (9.3 g)/(1.0 g/mol) = 9.3 mol.

mol O = mass/(atomic mass) = (36.2 g)/(16.0 g/mol) = 2.26 mol.

  • To get the empirical formula, we divide by the lowest no. of moles (2.26 mol) of O:

∴ C: H: O = (4.54 mol/2.26 mol) : (9.3 mol/2.26 mol) : (2.26 mol/2.26 mol) = 2: 4: 1.

<em>∴ Empirical formula mass of (C₂H₄O) = 2(atomic mass of C) + 4(atomic mass of H) + 1(atomic mass of O) =</em> 2(12.0 g/mol) + 4(1.0 g/mol) + (16.0 g/mol)<em> = 44.0 g/mol.</em>

∴ Number of times empirical mass goes into molecular mass = (88.0 g/mol)/(44.0 g/mol) = 2.0 times.

∴ The molecular formula is, 2(C₂H₄O), that is; <em>(C₄H₈O₂)</em>

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