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 of C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol
$$M(\text{CH}_3\text{OH}) = 12.01 + 4\times1.008 + 16.00 = 32.042\ \text{g/mol}$$

Step2: Find moles of CH₃OH

Moles = mass / molar mass
$$n = \frac{30.6\ \text{g}}{32.042\ \text{g/mol}} \approx 0.955\ \text{mol}$$

Step3: Calculate number of molecules

Use Avogadro's number $N_A = 6.022\times10^{23}\ \text{molecules/mol}$
$$N = n \times N_A = 0.955\ \text{mol} \times 6.022\times10^{23}\ \text{molecules/mol}$$

Answer:

Approximately $5.75\times10^{23}$ molecules