Question

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

b) Determine the percent composition (by mass) for each element in ammonium dichromate (NH4)2Cr2O7

c) A compound with a molecular mass of 202.23g/mol was found to have the following mass percent composition: 95.02% carbon and 4.98% hydrogen. Determine its:

*Empirical formula

*Molecular formula

Answer 1

Step-by-step explanation:

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

Number of phosphite ions in manganese (II) phosphite =
`7.23* 10^(22)`

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

Then
`7.23* 10^(22)` ions of phosphite will be ions will be in:

`(1)/(2)* 7.23* 10^(22)=1.446* 10^(23) molecules`

Moles of manganese (II) phosphite;=
`(1.446* 10^(23))/(6.022* 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:

`(2* 14g/mol)/(252.07 g/mol)* 100=11.10\%`

Percentage of hydrogen:

`(8* 1g/mol)/(252.07 g/mol)* 100=3.17\%`

Percentage of chromium:

`(2* 52 g/mol)/(252.07 g/mol)* 100=41.25\%`

Percentage of oxygen:

`(7* 16g/mol)/(252.07 g/mol)* 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 %

`(12 g/mol* x)/(202.23 g/mol)* 100=95.02\%`

x= 16.01 ≈ 16

Percentage of the hydrogen= 4.98 %

`(1 g/mol* y)/(202.23 g/mol)* 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_{(16)/(2)}H_{(10)/(2)}=C_8H_5`.