Question

1 gm of an organic compound gave 1.11 gm of magnesium pyrophosphate in the carius
method of estimation of phosphorus. % of P is
A) 11.11
B) 31
C) 22.2
D) 62
E) 33.3

Answer 1

Answer:

B)31

Explanation:

227 g of Mg2P2O7 (magnesium pyrophosphate) contains 62 g of P

Y g of Mg2P2O7 in W g of organic compound contains ( 62/222) x (y/w) of P

[here y is 1.11 and w is 1]

so % of Phosphorous = (62/222) x (1.11/1) x (100)

i.e 31%

Answer 2

Answer:

The percentage of phosphorus calculated is $31\%$.

Therefore, option b) $31\%$ is correct.

Explanation:

Given data,

The mass of the organic compound, $m_{1}$ = $1g$

The magnesium pyrophosphate's mass ( $Mg_{2}P_{2}O_{7}$ ), $m_{2}$ = $1.11g$

The percentage of phosphorus =?

By using the formula of % of P given below, we can find out the percentage of phosphorus:

• % of P = $\frac{Molar mass of P}{Molar mass of Mg_{2}P_{2}O_{7} } \times \frac{m_{2} }{m_{1} } \times 100$      -------equation (1)

Firstly, we have to calculate the value of molar masses.

• The molar mass of $Mg_{2}P_{2}O_{7}$ = 2 × atomic mass of Mg + 2 × atomic mass of P + 7 × atomic mass of O
• The molar mass of $Mg_{2}P_{2}O_{7}$ = $(2\times 24) + (2\times 31) + (7\times 16)$ = $48+ 62 + 112$ = $222g/mol$

And, the total number of Phosphorus present in $Mg_{2}P_{2}O_{7}$ = $2$

Therefore, the molar mass of $P_{2}$ = $2\times 31$ = $62g/mol$.

After substituting the calculated and the given values in equation (1), we get:

• % of P = $\frac{Molar mass of P}{Molar mass of Mg_{2}P_{2}O_{7} } \times \frac{m_{2} }{m_{1} } \times 100$  
• % of P = $\frac{62}{222 } \times \frac{1.11 }{1 } \times 100$
• % of P = $\frac{6882}{222}$
• % of P = $31\%$

Hence, the percentage of phosphorus estimated is $31\%$.