Question

part 3: multi-step conversions (grams ⇄ moles ⇄ particles)
problem
show work (2 pts)
ans (1 pt)
what is the mass of
1.204 × 10²⁴ molecules
of ethanol (c₂h₅oh)?
(molar mass = 46.07
g/mol)
steps: molecules →
moles → grams
how many carbon
atoms are in 25.0 g of
propane (c₃h₈)? (molar
mass = 44.11 g/mol)
steps: grams → moles
of c₃h₈ → moles of c →
atoms of c

Explanation:

Step1: Convert molecules to moles

Use Avogadro's number \(N_A = 6.022\times10^{23}\text{ molecules/mol}\).
The number of moles \(n\) of ethanol is \(n=\frac{1.204\times 10^{24}\text{ molecules}}{6.022\times 10^{23}\text{ molecules/mol}}\approx 2.0\text{ mol}\)

Step2: Convert moles to grams

Use the formula \(m = nM\), where \(M = 46.07\text{ g/mol}\) (molar mass of \(C_2H_5OH\))
\(m=(2.0\text{ mol})\times(46.07\text{ g/mol}) = 92.14\text{ g}\)

Step3: For the second problem (grams to moles of \(C_3H_8\))

Use \(n=\frac{m}{M}\), where \(m = 25.0\text{ g}\) and \(M = 44.11\text{ g/mol}\) (molar mass of \(C_3H_8\))
\(n=\frac{25.0\text{ g}}{44.11\text{ g/mol}}\approx0.567\text{ mol}\)

Step4: Moles of \(C\) in \(C_3H_8\)

Since 1 molecule of \(C_3H_8\) has 3 \(C\) atoms, moles of \(C\) is \(n_C=3\times0.567\text{ mol}=1.701\text{ mol}\)

Step5: Convert moles of \(C\) to atoms

Use \(N = nN_A\), where \(n = 1.701\text{ mol}\) and \(N_A=6.022\times 10^{23}\text{ atoms/mol}\)
\(N=(1.701\text{ mol})\times(6.022\times 10^{23}\text{ atoms/mol})\approx1.024\times 10^{24}\text{ atoms}\)

Answer:

For the first problem: \(92.14\text{ g}\)
For the second problem: \(1.024\times 10^{24}\text{ atoms}\)