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This is the second of our many quizzes on Percent composition. Click on this to access more such competitive quizzes on Physical Chemistry.

Percent composition Quiz 2

chemistry

 Find the percent composition of oxygen in sodium hydroxide?

  $\text{Percent composition of an element in a chemical} \\ = \dfrac{\text{Molecular weight of the respective element in the chemical}}{\text{Molecular weight of the total chemical}}\times 100$

  The molecular weight of the sodium hydroxide = 40. Molecular weight of the oxygen in sodium hydroxide is = 16. Percent composition of oxygen in NaOH= $\dfrac{\text{Molecular weight of the oxygen}}{\text{Molecular weight of the NaOH}}\times 100 \\  =\dfrac{16}{40}\times 100=40 \% $<br>

   Find the percent composition of copper in copper sulphate. The molecular weight of the copper sulphate is 159.60 g/mol.

 Molecular weight of copper = 63.55 $\\$ Molecular weight of copper sulphate is = 159.30 $\\$ Percent composition of copper in $CuS{{\text{O}}_{\text{4}}}\text{=}\dfrac{\text{Molecular weight of copper}}{\text{Molecular weight of the CuS}{{\text{O}}_{\text{4}}}}\times 100 \\  =\dfrac{63.55}{159.30}\times 100=39.81 \% $<br>

  Find the percent composition of each and every element in sulphuric acid. The molecular weight of the sulphuric acid is 98.079gm/mol. 

 $\text{Percent composition of an element in a chemical} \\ = \dfrac{\text{Molecular weight of the respective element in the chemical}}{\text{Molecular weight of the total chemical}}\times 100$

 Molecular weight of hydrogen = 1 $\\$ Molecular weight of sulphur = 32 $\\$ Molecular weight of oxygen atom = 16 $\\$ Percent composition of hydrogen=$ \\ \dfrac{\text{Molecular weight of hydrogen}}{\text{Molecular weight of sulphuric acid}}\times 100 \\=\dfrac{2}{98.07}\times 100=2.055 \%\ \\  \text{Percent composition of sulphur=} \\ \dfrac{\text{Molecular weight of sulphur}}{\text{Molecular weight of sulphuric acid}}\times 100 \\  =\dfrac{32}{98.07}\times 100=32.65 \%\ \\ \text{Percent composition of oxygen=} \\ \dfrac{\text{Molecular weight of oxygen}}{\text{Molecular weight of sulphuric acid}}\times 100 \\  =\dfrac{64}{98.07}\times 100=65.254 \%\ $

 Find the percent composition of all the elements involved in nitric acid. The molecular weight of the nitric acid is 63.01gm/mol.

  $\text{Percent composition of a element in a chemical} \\ = \dfrac{\text{Molecular weight of the respective element in the chemical}}{\text{Molecular weight of the total chemical}}\times 100$

 $ \text{Molecular weight of hydrogen = 1 }\\ \text{Molecular weight of nitrogen = 14 }\\ \text{Molecular weight of oxygen atom = 16}\\ \text{Percent composition of hydrogen=} \\ \dfrac{\text{Molecular weight of hydrogen}}{\text{Molecular weight of nitric acid}}\times 100 \\  =\dfrac{2}{63.01}\times 100=1.58 \%\ \\ \text{Percent composition of nitrogen =} \\ \dfrac{\text{Molecular weight of nitrogen}}{\text{Molecular weight of nitric acid}}\times 100 \\  =\dfrac{14}{63.01}\times 100=22.22 \%\ \\ \text{Percent composition of oxygen=} \\ \dfrac{\text{Molecular weight of oxygen}}{\text{Molecular weight of nitric acid}}\times 100 \\  =\dfrac{48}{63.01}\times 100=76.19 \%\ $

  Ethylene is an unsaturated hydrocarbon and has a chemical property to react with ozone and forms a product of carbonyl compounds. Find the percent composition of each element involved in the carbonyl compound.

 Ethylene reacts with ozone and forms formaldehyde as a product. 

$ \text{The molecular formula of formaldehyde is HCHO.}\\\text{The molecular weight of formaldehyde is 30.}\\ \text{Molecular weight of hydrogen = 1}\\ \text{Molecular weight of nitrogen = 14}\\ \text{Molecular weight of oxygen atom = 16}\\ \text{Percent composition of hydrogen=} \\ \\ \dfrac{\text{Molecular weight of hydrogen}}{\text{Molecular weight of formaldehyde}}\times 100 \\ =\dfrac{2}{30}\times 100=5.714\%\ \\ \text{Percent composition of carbon =} \\ \dfrac{\text{Molecular weight of carbon}}{\text{Molecular weight of formaldehyde}}\times 100 \\  =\dfrac{12}{30}\times 100=40.01 \%\ \\ \text{Percent composition of oxygen=} \\ \dfrac{\text{Molecular weight of oxygen}}{\text{Molecular weight of formaldehyde}}\times 100 \\  =\dfrac{16}{30}\times 100=54.285 \%\ $