Question

What is the mass, in grams, of 1.50 mol of iron (III) sulfate? Express your answer using three significant figures. How many moles of ammonium ions are in 6.935 g of ammonium carbonate? Express your answer using four significant figures. What is the mass, in grams, of 1.40 times 10^21 molecules of aspirin, C_9H_8O_4? Express your answer using three significant figures.

Answer 1

A. Mass = mole * molar mass

= 1.5 * 400

= 600 or 6.00*10^2 grams

B. Moles of ammonium carbonate = mass / molar mass

= 6.935 / 96

= 0.0722

(NH4)2CO3 = 2NH4+ + CO32-

Moles of ammonium ion = 2* 0.072 = 0.144

C. Moles of aspirin = number of molecules / avagadro number

= 1.40*10^21 / (6.022*10^23)

= 2.325*10^-3

Mass of aspirin = mole * molar mass

= 2.325*10^-3 * 180.158

= 0.419 grams

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