Question

Derive the equation for calculation of Percentage of Carbon and Hydrogen if mass of organic compound is W gm and mass of H2O or CO2 is m gm​

Answer 1

Given:-

• Mass of H2O and CO2 is m gm.
• Mass of organic compound is W gm.

For H

$\sf {H_{2}O(16 + 1 \times 2 \rightarrow18 \: gm) = 2H(2 \: gm)} \\ \\ \large{ \sf \: Here - } \\ \\ \rightarrow \sf{18 \: gm \: of \: H_{2}O \: contains \: 2 \: gm \: of \: hydrogen} \\ \\ \rightarrow \sf{{1 \: gm \: of \: H_{2}O \: contains \: \frac{2}{18} \: gm \: of \: hydrogen}} \\ \\ \large{ \sf{Therefore - }} \\ \\ \rightarrow \sf{{m \: gm \: of \: H_{2}O \: contains \: \frac{2}{18} \times m\: gm \: of \: hydrogen}}$

Percentage of hydrogen in W gm organic substance:-

$\red{\large \boxed{\% \: of \: H = \frac{2}{18} \times \frac{m}{W} \times 100}}$

For C

$\sf{CO_{2}(16 \times 2 + 12 \rightarrow 44gm}) = C(12gm) \\ \\ \large{\sf{Here - }} \\ \\ \rightarrow \sf44 \: gm \: of \: CO_{2} \: contains \: 12 \: gm \: carbon \\ \\ \rightarrow \sf1 \: gm \: of \: CO_{2} \: contains \: \frac{12}{44} \: gm \: carbon \\ \\ \large{ \sf{Therefore - }} \\ \\ \rightarrow \sf \: m \: gm \: of \: CO_{2} \: contains \: \frac{12}{44} \times m \: gm \: carbon$

Percentage of C in W gm organic substance:-

$\blue{ \large \sf{\boxed{\% \: of \: C = \frac{12}{44} \times \frac{m}{W} \times 100}}}$

Answer 2

Answer:

Mass of H2O and CO2 is m gm.

Mass of organic compound is W gm.

For H

\begin{gathered}\sf {H_{2}O(16 + 1 \times 2 \rightarrow18 \: gm) = 2H(2 \: gm)} \\ \\ \large{ \sf \: Here - } \\ \\ \rightarrow \sf{18 \: gm \: of \: H_{2}O \: contains \: 2 \: gm \: of \: hydrogen} \\ \\ \rightarrow \sf{{1 \: gm \: of \: H_{2}O \: contains \: \frac{2}{18} \: gm \: of \: hydrogen}} \\ \\ \large{ \sf{Therefore - }} \\ \\ \rightarrow \sf{{m \: gm \: of \: H_{2}O \: contains \: \frac{2}{18} \times m\: gm \: of \: hydrogen}}\end{gathered}

H

2

O(16+1×2→18gm)=2H(2gm)

Here−

→18gmofH

2

Ocontains2gmofhydrogen

→1gmofH

2

Ocontains

18

2

gmofhydrogen

Therefore−

→mgmofH

2

Ocontains

18

2

×mgmofhydrogen

Percentage of hydrogen in W gm organic substance:-

\red{\large \boxed{\% \: of \: H = \frac{2}{18} \times \frac{m}{W} \times 100}}

%ofH=

18

2

×

W

m

×100

For C

\begin{gathered}\sf{CO_{2}(16 \times 2 + 12 \rightarrow 44gm}) = C(12gm) \\ \\ \large{\sf{Here - }} \\ \\ \rightarrow \sf44 \: gm \: of \: CO_{2} \: contains \: 12 \: gm \: carbon \\ \\ \rightarrow \sf1 \: gm \: of \: CO_{2} \: contains \: \frac{12}{44} \: gm \: carbon \\ \\ \large{ \sf{Therefore - }} \\ \\ \rightarrow \sf \: m \: gm \: of \: CO_{2} \: contains \: \frac{12}{44} \times m \: gm \: carbon\end{gathered}

CO

2

(16×2+12→44gm)=C(12gm)

Here−

→44gmofCO

2

contains12gmcarbon

→1gmofCO

2

contains

44

12

gmcarbon

Therefore−

→mgmofCO

2

contains

44

12

×mgmcarbon

Percentage of C in W gm organic substance:-

\blue{ \large \sf{\boxed{\% \: of \: C = \frac{12}{44} \times \frac{m}{W} \times 100}}}

%ofC=

44

12

×

W

m

×100

Explanation:

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