Chemistry, 25.07.2019 03:10 paytonpaige22

When 0.6943 g of a compound is subjected to combustion analysis it produced 1.471 g co2 and 0.391 g h2o. what is its empirical and molecular formula if its molar mass is 172 g/mol if the compound is composed of only carbon, hydrogen, and oxygen.

Answers: 2

Show answers

Answers

Answer from: preciosakassidy

The empirical molecular formula for the given organic compound are C_2H_3O and $C_8H_{12}O_4$ respectively.

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.

We are given:

Mass of CO_2=1.471g

Mass of H_2O=0.391g

We know that:

Molar mass of carbon dioxide = 44 g/mol

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 1.471 g of carbon dioxide, $\frac{12}{44}\times 1.471=0.401g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.391 g of water, $\frac{2}{18}\times 0.391=0.0434g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.6943) - (0.401 + 0.0434) = 0.2499 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.401g}{12g/mole}=0.0334moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0434g}{1g/mole}=0.0434moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2499g}{16g/mole}=0.0156moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0156 moles.

For Carbon = $\frac{0.0334}{0.0156}=2.14\approx 2$

For Hydrogen = $\frac{0.0434}{0.0156}=2.78\approx 3$

For Oxygen = $\frac{0.0156}{0.0156}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_{3}O_1=C_2H_3O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 172 g/mol

Mass of empirical formula = 43 g/mol

Putting values in above equation, we get:

$n=\frac{172g/mol}{43g/mol}=4$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_8H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound are C_2H_3O and $C_8H_{12}O_4$ respectively.

Answer from: Quest

answer:

technetium is the most reactive. this is followed by silver. the least reactive out of these three elements is strontium.

Answer from: Quest

1. i don't know.2. bitter taste, feels slippery, contains hydroxide or produces it in a solution.3. 104.i don't know.5. electrons.hope this somewhat!

Another question on Chemistry

Chemistry, 22.06.2019 00:30

What must happen before a body cell can begin mitotic cell division

Answers: 1

Answer

Chemistry, 22.06.2019 05:00

Use the table to identify the phase and phase changes of the elements under the given conditions. write the name of the substance, phase, or phase change

Answers: 3

Answer

Chemistry, 22.06.2019 16:30

How do i turn 2 s into one in the products

Answers: 1

Answer

Chemistry, 22.06.2019 18:00

What amount of heat is exchanged when 106.2 grams of substance y goes from a liquid at 35 degrees celsius to a solid at the same temperature? melting point of substance y = 35 degrees c; δhvaporization = 3.67 j/mol; δhfusion = 3.30 j/mol. mwsubstance y = 28.22 g/mol. −12.4 j −3.51 x 102 j 1.24 x 101 j 351 j

Answers: 1

Answer

You know the right answer?

When 0.6943 g of a compound is subjected to combustion analysis it produced 1.471 g co2 and 0.391 g...

Questions