Question

calculate the percentage composition of the following​

Answer 1

$\large \boxed{1)- C_6H_12O_6}$

Atomic weight of

▶Carbon C = 12g

▶Hydrogen H = 1g

▶Oxygen O = 16 g

Molecular weight = 6(C) + 12(H) + 6(O)

= 6 × 14 + 12 ×1 + 6 × 16

=84 + 12 + 96

=192g

◼For percentage of carbon in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{84}{192} \times 100}$

$\large \boxed{Percentage \: composition =6</p><p>43.75\%}$

◼For percentage of hydrogen in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{12}{192} \times 100}$

$\large \boxed{Percentage \: composition =6.25\%}$

◼For percentage of oxygen in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{96}{192} \times 100}$

$\large \boxed{Percentage \: composition =50\%}$

$\large \boxed{2)- CO_2}$

Atomic weight of

▶Carbon C = 12g

▶Oxygen O = 16 g

Molecular weight = (C) + 2(O)

= 12 + 2(16)

=12 + 32

=44g

◼For percentage of carbon in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{12}{44} \times 100}$

$\large \boxed{Percentage \: composition =27.27\%}$

◼For percentage of oxygen in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{32}{44} \times 100}$

$\large \boxed{Percentage \: composition =72.72\%}$

$\large \boxed{3)- CO}$

Atomic weight of

▶Carbon C = 12g

▶Oxygen O = 16 g

Molecular weight = (C) + (O)

= 12 + 16

=28 g

◼For percentage of carbon in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{12}{28} \times 100}$

$\large \boxed{Percentage \: composition =42.85\%}$

◼For percentage of oxygen in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{16}{28} \times 100}$

$\large \boxed{Percentage \: composition =57.14\%}$

$\large \boxed{2)- NO_2}$

Atomic weight of

▶Nitrogen = 14g

▶Oxygen O = 16 g

Molecular weight = (14) + 2(O)

= 14 + 2(16)

=14 + 32

=46g

◼For percentage of carbon in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{12}{46} \times 100}$

$\large \boxed{Percentage \: composition =30.43\%}$

◼For percentage of oxygen in compound,

$\boxed{Percentage \: composition = \dfrac{WEIGHT \: OF\: ONE \: MOLECULE \: COMPOUND}{GRAM \: MOLECULAR \: WEIGHT \: OF \: THE \:COMPOUND} \times 100}$

$\mathsf{Percentage \: composition = \dfrac{32}{46} \times 100}$

$\large \boxed{Percentage \: composition =69.56\%}$