13. An and oxygen atoms. nalze th oloing data for a 5.530 g sample ofa compound

Question

13. An and oxygen atoms. nalze th oloing data for a 5.530 g sample ofa compound that contains only carbon, hydrogen alyze the following data for a 5.530 g sample of a compound that contains only carbon, hydrogen (a) If the sample contains 1.630 x 1023 atoms of carbon, how many moles of carbon are present in the sample? (b) 5.530 g of the compound contains 0.5458 g of hydrogen atoms. How many moles of hydrogen are present in the sample? (c) The sample is 31.33 % oxygen by mass. Calculate the mass of oxygen (in grams) present in the sample (d) What is the empirical formula of the compound?

Explanation / Answer

a) 1 mole contains 6.023 * 10^(23) atoms of carbon

Number of moles of Carbon = 1.630 * 10^(23)/6.023 * 10^(23) = 0.27062 moles

b) Molar mass of H2 = 2 gm/mol

Number of moles of Hydrogen gas = 0.5458/2 = 0.2729 moles

Moles of Hydrogen = 0.5458 moles

c) Mass of Oxygen in the sample = 31.33/100 * 5.530 = 1.7325 gms

d) Empirical formula of the compound

number of moles of oxygen = 1.7325/32 = 0.054142 moles

Hence the empirical formula will be C2H4O

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