Question

d. if 30.6 g of methanol (ch₃oh) is used to start a fire, how many molecules were used?

Explanation:

Step1: Calculate molar mass of CH₃OH

Molar mass = $12.01 + (4\times1.008) + 16.00$ = $32.042\ \text{g/mol}$

Step2: Find moles of CH₃OH

Moles = $\frac{\text{Mass}}{\text{Molar Mass}}$ = $\frac{30.6}{32.042} \approx 0.955\ \text{mol}$

Step3: Calculate number of molecules

Molecules = Moles $\times N_A$, where $N_A = 6.022\times10^{23}\ \text{molecules/mol}$
Expression: $0.955 \times 6.022\times10^{23} \approx 5.75\times10^{23}$

Answer:

Approximately $5.75\times10^{23}$ molecules of methanol were used.