Explanation:
a) Manganese (II) phosphite :![Mn_3(PO_3)_2](https://tex.z-dn.net/?f=Mn_3%28PO_3%29_2)
Number of phosphite ions in manganese (II) phosphite = ![7.23\times 10^{22}](https://tex.z-dn.net/?f=7.23%5Ctimes%2010%5E%7B22%7D)
In one molecule of manganese (II) phosphite there are 2 ions of phosphite.
Then
ions of phosphite will be ions will be in:
![\frac{1}{2}\times 7.23\times 10^{22}=1.446\times 10^{23} molecules](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Ctimes%207.23%5Ctimes%2010%5E%7B22%7D%3D1.446%5Ctimes%2010%5E%7B23%7D%20molecules)
Moles of manganese (II) phosphite;= ![\frac{1.446\times 10^{23}}{6.022\times 10^{23}}= 0.2401 mol](https://tex.z-dn.net/?f=%5Cfrac%7B1.446%5Ctimes%2010%5E%7B23%7D%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%7D%3D%200.2401%20mol)
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\%](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Ctimes%2014g%2Fmol%7D%7B252.07%20g%2Fmol%7D%5Ctimes%20100%3D11.10%5C%25)
Percentage of hydrogen:
![\frac{8\times 1g/mol}{252.07 g/mol}\times 100=3.17\%](https://tex.z-dn.net/?f=%5Cfrac%7B8%5Ctimes%201g%2Fmol%7D%7B252.07%20g%2Fmol%7D%5Ctimes%20100%3D3.17%5C%25)
Percentage of chromium:
![\frac{2\times 52 g/mol}{252.07 g/mol}\times 100=41.25\%](https://tex.z-dn.net/?f=%5Cfrac%7B2%5Ctimes%2052%20g%2Fmol%7D%7B252.07%20g%2Fmol%7D%5Ctimes%20100%3D41.25%5C%25)
Percentage of oxygen:
![\frac{7\times 16g/mol}{252.07 g/mol}\times 100=44.44\%](https://tex.z-dn.net/?f=%5Cfrac%7B7%5Ctimes%2016g%2Fmol%7D%7B252.07%20g%2Fmol%7D%5Ctimes%20100%3D44.44%5C%25)
c) Molar mass of the substance = 202.23 g/mol
Percentage of the hydrogen = 4.98 %
Let the molecular formula be ![C_xH_y](https://tex.z-dn.net/?f=C_xH_y)
Percentage of the carbon = 95.02 %
![\frac{12 g/mol\times x}{202.23 g/mol}\times 100=95.02\%](https://tex.z-dn.net/?f=%5Cfrac%7B12%20g%2Fmol%5Ctimes%20x%7D%7B202.23%20g%2Fmol%7D%5Ctimes%20100%3D95.02%5C%25)
x= 16.01 ≈ 16
Percentage of the hydrogen= 4.98 %
![\frac{1 g/mol\times y}{202.23 g/mol}\times 100=4.98\%](https://tex.z-dn.net/?f=%5Cfrac%7B1%20g%2Fmol%5Ctimes%20y%7D%7B202.23%20g%2Fmol%7D%5Ctimes%20100%3D4.98%5C%25)
y= 10.07 ≈ 10
The molecular formula of the substance is
.
The empirical formula of the substance is
.