Respuesta :
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:
479.36 grams is the mass (in grams) of [tex]9.02\times 10^{24}[/tex] molecules of methanol .
Explanation:
[tex]n=\frac{m}{M}[/tex]
[tex]N=n\times N_A[/tex]
n = Moles of compound
m = mass of the compound
M = molar mass of the compound
N = Number of molecules or atoms of compounds
[tex]N_A=6.022\times 10^{23} mol^{-1}[/tex]
Given :
Number of methanol molecules = N
[tex]N=9.02\times 10^{24}[/tex] molecules
n = ?
[tex]N=n\times N_A[/tex]
[tex]n=\frac{N}{N_A}=\frac{9.02\times 10^{24}}{6.022\times 10^{23} mol^{-1}}=14.98 mol[/tex]
Molar mass of methanol = M = 32 g/mol
m = ?
[tex]n=\frac{m}{M}[/tex]
[tex]m=n\times M=14.98 mol\times 32 g/mol=479.36 g[/tex]
479.36 grams is the mass (in grams) of [tex]9.02\times 10^{24}[/tex] molecules of methanol .