What is the percentage composition of #(NH_4)_2CO_3#?

1 Answer
Feb 28, 2016

Answer:

#"29.155% N"#
#"8.392% H"#
#"12.500% C"#
#"49.952% O"#

Explanation:

In order to determine percent composition of the elements in a compound, the molar mass (MM) of each element is divided by the MM of the compound and multiplied by 100.

#"% composition of element"=("MM of element")/("MM of compound")xx100#

The molar mass of ammonium carbonate #("(NH"_4")"_2"CO"_3")# is #"96.086 g/mol"#. (Determined by multiplying the subscripts of each element times its MM using its atomic mass in g/mol on the periodic table.)

Determine the MM of each element in the compound by multiplying the subscript of each element by its MM (relative atomic mass in g/mol from the periodic table).

#"N":# #(2xx14.007"g/mol")="28.014 g N"#
#"H":# #(8xx1.008"g/mol")="8.064 g H"#
#"C":# #(1xx12.011"g/mol")="12.011 g C"#
#"O":# #(3xx15.999"g/mol")="47.997 g O"#

Now determine the percent composition of each element in the compound.

#%"composition N"=("28.014 g/mol")/("96.086 g/mol")xx100="29.155% N"#

#%"composition H"=("8.064 g/mol")/("96.086 g/mol")xx100="8.392% H"#

#%"composition C"=("12.011 g/mol")/("96.086 g/mol")xx100="12.500% C"#

#%"composition O"=("47.997 g/mol")/("96.086 g/mol")xx100="49.952% O"#

Add the percentages: #29.155%+8.392%+12.500%+49.952%="99.999%"~~100%"#