Question

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

Answer 1

479.36 grams is the mass (in grams) of
`9.02* 10^(24)` molecules of methanol .

Step-by-step explanation:

`n=(m)/(M)`

`N=n* 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* 10^(23) mol^(-1)`

Given :

Number of methanol molecules = N

`N=9.02* 10^(24)` molecules

n = ?

`N=n* N_A`

`n=(N)/(N_A)=(9.02* 10^(24))/(6.022* 10^(23) mol^(-1))=14.98 mol`

Molar mass of methanol = M = 32 g/mol

m = ?

`n=(m)/(M)`

`m=n* M=14.98 mol* 32 g/mol=479.36 g`

479.36 grams is the mass (in grams) of
`9.02* 10^(24)` molecules of methanol .

Answer 2

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.