Question

calculate the percent composition of :1) iron in ferric oxide ,Fe2O3(atomic mass of Fe=56 2) nitrogen in urea,H2NCNH2

Answer 1

fe%=112/112+48*100= 11200/160=70%
n%= 2800/28+4+12= 2800/44=63.6%

Answer 2

Calculation of percent composition

1. Iron in $Fe_2O_3$:

• Mass% = $\frac{Mass}{Total mass}\times 100$
• Total mass of given substance is

= 2( Atomic weight of Fe) + 3 (Atomic weight of O)

= 2(56)+ 3(16)

= 112+ 48

= 160.

• The total mass of the compound is 160, and the mass of iron is 112.
• Thus mass percentage of Fe is

Mass % of Fe = $\frac{Mass}{Total mass}\times 100$

Mass % of Fe = $\frac{112}{160}\times 100$

Mass % of Fe = 70%

• Hence the percent composition of iron in $Fe_2O_3$ is 70%.

2. Nitrogen in $CH_4N_2O$:

• Total mass of given substance is

= 1 (At. mass of C) + 4(At. mass of H) + 2 (At. mass of N) + 1 (At. mass of O)

= 1 (12) + 4 (1) + 2 (14) + 1 (16)

= 12 + 4 + 28 + 16

= 60

• The total mass of the compound is 60, and the mass of nitrogen is 28.
• Thus mass percentage of N is

Mass % of N = $\frac{Mass}{Total mass}\times 100$

Mass % of N = $\frac{28}{60}\times 100$

Mass % of N = 47%

• Hence the percent composition of nitrogen in $CH_4N_2O$  is 47%.

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