Mole Calculations (1 Viewer)

[hunterbear]

Hi!

I would really appreciate some help on this question from the 2013 Chemistry Science Olympiad Exam.

If a chemist requires 16.0 mole of liquid ethanol (C2H5OH) for a chemical reaction, what volume should be used? The density of C2H5OH is 0.789 g mL–1.

 

[InteGrand]

Hi!

I would really appreciate some help on this question from the 2013 Chemistry Science Olympiad Exam.

If a chemist requires 16.0 mole of liquid ethanol (C2H5OH) for a chemical reaction, what volume should be used? The density of C2H5OH is 0.789 g mL–1.

Find how many grams are needed (i.e. how many grams in 16 mol), using n = m/M, i.e. mass required = (number of moles required)×(molar mass).

Then: volume required = mass required ÷ density.

 

[InteGrand]

Molar mass of ethanol is M = 46.06844 g/mol.

Hence the mass required is m_required = n_required × M

i.e. m_required = 16.0 mol × 46.06844 g/mol ≈ 737.09504 g.

So the required volume is V_required = m_required ÷ ρ_ethanol, where ρ_ethanol is the density of ethanol, given to be ρ_ethanol = 0.789 g mL^-1.

⇒ V_required ≈ 737.09504 g ÷ 0.789 g mL^-1 ≈ 934.214246 mL ≈ 934 mL (to 3 significant figures).

∴ About 934 mL of ethanol is required.