108th Problem 2016

Chemistry Level 2

What is the Empirical Formula \text{Empirical Formula} of a compound containing the following?:

  • 43.38 % N a 43.38 \text{\% }\ce{Na}
  • 11.33 % C 11.33 \text{\% }\ce{C}
  • 45.29 % O 45.29 \text{\% }\ce{O}

Assume the weight of the sample is 100 grams. 100 \text{ grams.}

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N X a X 3 C O X 2 \ce{N_a_3CO2} N X a X 3 C O X 4 \ce{N_a_3CO4} N X a X 2 C O X 3 \ce{N_a_2CO3} N X a X 4 C O X 4 \ce{N_a_4CO4}

Angela Fajardo by Angela Fajardo , 19, Philippines

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2 solutions

Ashish Menon
Apr 25, 2016

Relevant wiki: Stoichiometry

Element x = % composition y = Mol. weight x/y Simplest Ratio
Na 43.38 23 43.38 ÷ 23 = 1.88 1.88 ÷ 0.94 = 2
C 11.33 12 11.33 ÷ 12 = 0.94 0.94÷ 0.94 = 1
O 45.29 16 45.29 ÷ 16 = 2.82 2.82 ÷ 0.94 = 3

∴ \therefore The empirical formula is N a 2 C O 3 \boxed{{Na}_2CO_3}

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Abhiram Rao
Apr 26, 2016

Direct application of this method. Divide percentages by their molecular mass and then divide it by the least ratio of percentages and mass and then simplest whole number ratio. So it's Na2CO3

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