Question

What is the mass (in grams) of 9.02 × 1024 molecules of methanol (CH3OH)?

Answer 1

You first calculate the formula mass of methanol (CH3OH). This is, add the mass of each element in the formula. C: 12 g/mol; H: 4* 1 g/mol = 4 g/mol; O: 16 g/mol => formula mass = 12 g/mol + 4g/mol + 16 g/mol = 32 g/mol. Second, calculate the number of moles in 9,02 * 10^24 molecules, using Avogadros number: number of moles = 9.02 * 10^24 molecules / 6.02 * 10^23 molecules/mol = 15 moles. Third, the mass in grams of the compound equals the number of moles times the formula mass = 15 moles * 32 g/mol = 480 g. Answer: 480 grams.

Answer 2

Answer:

479.36 grams is the mass (in grams) of $9.02\times 10^{24}$  molecules of methanol .

Explanation:

$n=\frac{m}{M}$

$N=n\times N_A$

n = Moles of compound

m = mass of the compound

M = molar mass of the compound

N = Number of molecules or atoms of compounds

$N_A=6.022\times 10^{23} mol^{-1}$

Given :

Number of methanol molecules = N

$N=9.02\times 10^{24}$ molecules

n = ?

$N=n\times N_A$

$n=\frac{N}{N_A}=\frac{9.02\times 10^{24}}{6.022\times 10^{23} mol^{-1}}=14.98 mol$

Molar mass of methanol = M = 32 g/mol

m = ?

$n=\frac{m}{M}$

$m=n\times M=14.98 mol\times 32 g/mol=479.36 g$

479.36 grams is the mass (in grams) of $9.02\times 10^{24}$  molecules of methanol .