Question

A) How many moles of ammonium ions are in 0.732 g of ammonium carbonate?

B) What is the mass, in grams, of 0.0331 mol of iron(III) phosphate?

C) What is the mass, in grams, of 1.70 x10^23 molecules of aspirin, C9H8O4?

D) What is the molar mass of a particular compound of 0.086 mol weighs 6.98 g?

Answer 1

Solution:

A) $moles=\dfrac{m}{M}=\dfrac{0.792g}{96(\frac{g}{mole})}=8.25*10^{-3}moles$

B) $mass=moles*M=0.0331 \;moles*\; 151\;\frac{g}{moles}=4.9981g$

C) $\dfrac{molecules}{N_A}=\dfrac{1.7*10^{23}}{6.02*10^{23}}=0.282moles$

$mass=moles*M=0.282\;moles*\;180\frac{g}{moles}=50.76g$

D) $M=\dfrac{mass}{moles}=\dfrac{6.98g}{0.086moles}=81.12 \;(\frac{g}{moles})$