Question

A) What is the mass in grams of a sample of manganese (II) phosphite containing 7.23 x 10^22 phosphite ions?

Answer 1

Explanation:

a) Manganese (II) phosphite :$Mn_3(PO_3)_2$

Number of phosphite ions in manganese (II) phosphite = $7.23\times 10^{22}$

In one molecule of manganese (II) phosphite there are 2 ions of phosphite.

Then $7.23\times 10^{22}$ ions of phosphite will be ions will be in:

$\frac{1}{2}\times 7.23\times 10^{22}=1.446\times 10^{23} molecules$

Moles of manganese (II) phosphite;= $\frac{1.446\times 10^{23}}{6.022\times 10^{23}}= 0.2401 mol$

Mass of 0.2401 mol of Manganese (II) phosphite :

0.2401 mol × 322.75 g/mol = 77.49 g

77.49 is the mass in grams of a sample of manganese (II) phosphite.

b) Molar mass of ammonium chromate =252.07 g/mol

Percentage of Nitrogen:

$\frac{2\times 14g/mol}{252.07 g/mol}\times 100=11.10\%$

Percentage of hydrogen:

$\frac{8\times 1g/mol}{252.07 g/mol}\times 100=3.17\%$

Percentage of chromium:

$\frac{2\times 52 g/mol}{252.07 g/mol}\times 100=41.25\%$

Percentage of oxygen:

$\frac{7\times 16g/mol}{252.07 g/mol}\times 100=44.44\%$

c) Molar mass of the substance = 202.23 g/mol

Percentage of the hydrogen = 4.98 %

Let the molecular formula be $C_xH_y$

Percentage of the carbon = 95.02 %

$\frac{12 g/mol\times x}{202.23 g/mol}\times 100=95.02\%$

x= 16.01 ≈ 16

Percentage of the hydrogen= 4.98 %

$\frac{1 g/mol\times y}{202.23 g/mol}\times 100=4.98\%$

y= 10.07 ≈ 10

The molecular formula of the substance is $C_{16}H_{10}$.

The empirical formula of the substance is $C_{\frac{16}{2}}H_{\frac{10}{2}}=C_8H_5$.