A sample of a compound contains #"0.672 g Co"#, #"0.569 g As"#, and #"0.486 g O"#. What is its empirical formula?

2 Answers

The empirical formula is #"Co"_3"As"_2"O"_8#.

Explanation:

#color(white)(XX)"Atomic"#
#color(white)(XX)"Weight"color(white)(X)"Weight (g)"color(white)(X)"Wt./At. wt."color(white)(l)"Ratio"color(white)(Xll)"×2"#
#"Co"color(white)(Xl)58.9 color(white)(XXX)0.672color(white)(X)color(white)(XXX)0.0114color(white)(XXl)1.50color(white)(XX)3.00#
#"As"color(white)(Xl)74.9color(white)(XXX)0.569color(white)(XXXX)0.00760color(white)(XX)1color(white)(XXXl)2#
#"O"color(white)(Xll)16.0color(white)(XXX)0.486color(white)(XXXX)0.0304color(white)(XXll)4.00color(white)(XX)8.00#

First we have to calculate the ratio of the weight of each element to its atomic weight.

#"Co: 0.672/58.9 = 0.0114"#
#"As: 0.569/74.9 = 0.00760"#
#"O: 0.486/16.0 =" color(white)(ll)0.0304#

After calculating the ratio, find lowest value for the element, i.e. #0.00760#.

Divide this number into all the values above.

#"Co = 0.0114/0.0076 = 1.50"#
#"As = 0.0076/0.0076 = 1"#
#"O = 0.0304/0.0076 ="color(white)(ll)4.00#

Since the ratio for #"Co"# is not an integer, all ratios are multiplied by 2, which gives the empirical formula #"Co"_3"As"_2"O"_8#.

Sep 30, 2015

The empirical formula is #"Co"_3"As"_2"O"_8"# .

Explanation:

#0.672"g Co"#
#0.569"g As"#
#0.486"g O"#

Molar mass of each element (atomic weight in grams/mole)

#"Co":# #58.9332"g/mol"#
#"As":# #74.9216"g/mol"#
#"O":# #15.999"g/mol"#

Determine the number of moles of each element using the given mass and the molar mass.

#0.672"g Co"xx(1"mol Co")/(58.9332"g Co")="0.01152 mol Co"#

#0.569"g As"xx(1"mol As")/(74.9216"g As")="0.007594 mol As"#

#0.486"g O"xx(1"mol O")/(15.999"g O")="0.03038 mol O"#

Moles Ratios and Empirical Formula

Determine the mole ratios by dividing the number of moles of each element by the least number of moles.

#(0.01152)/(0.007594)"mol Co"=1.517#
#(0.007594)/(0.007594)"mol As"=1.00#
#(0.03038)/(0.007594)"mol O"=4.00#

An empirical formula is the lowest whole number ratio of elements in the compound. Problem: the mole ratio of Co is 1.5, which is not a whole number. In this case you multiply each ratio times 2.

The empirical formula is #"Co"_3"As"_2"O"_8"#.