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Ostrovityanka [42]
3 years ago
11

Silver was precipitated and collected by filtration. The experiment yielded .077g of silver after dying. The predicted yield was

.083g. The percent error present in the collection would be ?
Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:

Precentage error = 7.23%

Explanation:

The precentage error formula  is the  useful method  for determining precision of your experimental result

It can be calculated using :

percent\ error = |\frac{experimental\ value-predicted\ value} {predicted\ value}|\times 100

percent\ error = |\frac{0.077\ -0.083} {0.083}|\times 100

So we get percentage error =  7.23%

Precision means how close the experimental value to the true value or theoretical value

If percentage error is large it means experimental results are deviated and there is need to check mistakes,imprecision in the experiment

If the percentage error is zero it means the experiment is perfectly done without  inaccuracy

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Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

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Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, =\frac{12}{44}\times 12.24=3.338g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, =\frac{2}{18}\times 2.505=0.278g of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.278}{0.104}=3

For H =\frac{0.278}{0.104}=3

For O =\frac{0.104}{0.104}=1

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is C_3H_3O.

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3 years ago
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