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How many molecules of ammonium phosphate (NH4)3PO4 are in 2.00 grams sample of it?

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To solve this problem, we need to use the concept of Avogadro's number and the molar mass of the compound.

The molar mass of (NH4)3PO4 can be calculated by adding the molar mass of its individual elements:

M(N) = 14.01 g/mol

M(H) = 1.01 g/mol

M(P) = 30.97 g/mol

M(O) = 16.00 g/mol

Molar mass of (NH4)3PO4 = 3(M(N) + 4M(H)) + M(P) + 4(M(O))

= 3(14.01 g/mol + 4(1.01 g/mol)) + 30.97 g/mol + 4(16.00 g/mol)

= 149.09 g/mol

Now, we can calculate the number of moles of (NH4)3PO4 in 2.00 g sample:

Number of moles = mass ÷ molar mass

= 2.00 g ÷ 149.09 g/mol

= 0.0134 mol

Finally, we can use Avogadro's number to convert the number of moles to the number of molecules:

Number of molecules = number of moles × Avogadro's number

= 0.0134 mol × 6.022 × 10^23 molecules/mol

= 8.06 × 10^21 molecules

Therefore, there are 8.06 × 10^21 molecules of (NH4)3PO4 in a 2.00 g sample of it.

User Pavel Kostenko
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