For a compound NaAc, the percent composition is that: 29.28% C, 3.68% H, 39.01% O, and 28.03% Na. What is the empirical formula of this compound?

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Answer 1 · 2017-04-05T18:32:14

#"NaC"_2"H"_3"O"_2#

Convert the percentages to grams. If there were 100 g, each 1 % would equal 1 g.

So,

#"29.28 % C = 29.28g C"#
#"3.68 % H = 3.68 g H"#
#"39.01% O" = "39.01 g O"#
#"28.03% Na" = "28.03 g Na"#

Then convert the grams to moles

# ("29.28 g")/("12.01 g/mol") = "2.438 mol C"#

#("3.68 g")/ ("1.008 g/mol") = "3.65 mol H"#

#("39.01 g")/("16.00 g/mol") = "2.438 mol O"#

# ("28.03 g")/("22.99 g/mol")= "1.219 mol Na"#

Convert these numbers of moles to mole ratios by dividing all of them by the smallest mole number.

# 2.438 /1.219 = 2.000/1 = "2 O: 1 Na"#

# 2.438/ 1.219= 2.000/1 = "2 C : 1 Na"#

# 3.65/1.219 = 2.99/1 = "3 H : 1 Na"#

So, the molar ratio is #"1 Na : 2 C : 2 O : 3 H"#,

and the empirical formula is

#"NaC"_2"H"_3"O"_2#