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

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

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Answer 1 · 2015-09-30T14:36:00

The empirical formula is ${Co}_{3} {As}_{2} O_{8}$ $"Co"_3"As"_2"O"_8$ .

Explanation:

$X X Atomic$ $color(white)(XX)"Atomic"$
$X X Weight X Weight (g) X Wt./At. wt. l Ratio X l l \times2$ $color(white)(XX)"Weight"color(white)(X)"Weight (g)"color(white)(X)"Wt./At. wt."color(white)(l)"Ratio"color(white)(Xll)"×2"$
$Co X l 58.9 X X X 0.672 X X X X 0.0114 X X l 1.50 X X 3.00$ $"Co"color(white)(Xl)58.9 color(white)(XXX)0.672color(white)(X)color(white)(XXX)0.0114color(white)(XXl)1.50color(white)(XX)3.00$
$As X l 74.9 X X X 0.569 X X X X 0.00760 X X 1 X X X l 2$ $"As"color(white)(Xl)74.9color(white)(XXX)0.569color(white)(XXXX)0.00760color(white)(XX)1color(white)(XXXl)2$
$O X l l 16.0 X X X 0.486 X X X X 0.0304 X X l l 4.00 X X 8.00$ $"O"color(white)(Xll)16.0color(white)(XXX)0.486color(white)(XXXX)0.0304color(white)(XXll)4.00color(white)(XX)8.00$

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

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

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

Divide this number into all the values above.

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

Since the ratio for $Co$ $"Co"$ is not an integer, all ratios are multiplied by 2, which gives the empirical formula ${Co}_{3} {As}_{2} O_{8}$ $"Co"_3"As"_2"O"_8$ .

Answer 2 · 2015-09-30T17:23:45

The empirical formula is ${Co}_{3} {As}_{2} O_{8}$ $"Co"_3"As"_2"O"_8"$ .

Explanation:

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

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

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

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

$0.672 g Co \times \frac{1 mol Co}{58.9332 g Co} = 0.01152 mol Co$ $0.672"g Co"xx(1"mol Co")/(58.9332"g Co")="0.01152 mol Co"$

$0.569 g As \times \frac{1 mol As}{74.9216 g As} = 0.007594 mol As$ $0.569"g As"xx(1"mol As")/(74.9216"g As")="0.007594 mol As"$

$0.486 g O \times \frac{1 mol O}{15.999 g O} = 0.03038 mol O$ $0.486"g O"xx(1"mol O")/(15.999"g O")="0.03038 mol O"$

Moles Ratios and Empirical Formula

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

$\frac{0.01152}{0.007594} mol Co = 1.517$ $(0.01152)/(0.007594)"mol Co"=1.517$
$\frac{0.007594}{0.007594} mol As = 1.00$ $(0.007594)/(0.007594)"mol As"=1.00$
$\frac{0.03038}{0.007594} mol O = 4.00$ $(0.03038)/(0.007594)"mol O"=4.00$

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 ${Co}_{3} {As}_{2} O_{8}$ $"Co"_3"As"_2"O"_8"$ .

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

2 Answers

Explanation:

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... and beyond

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

2 Answers

Explanation:

Explanation:

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