Chemistry, 01.07.2019 21:20 robert7248

A21.82 gram sample of an organic compound containing c, h and o is analyzed by combustion analysis and 21.33 grams of co2 and 4.366 grams of h2o are produced. in a separate experiment, the molar mass is found to be 90.04 g/mol. determine the empirical formula and the molecular formula of the organic compound.

Answers: 2

Show answers

Answers

Answer from: lin550

The molecular formula for the given organic compound is C_2H_2O_4

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=21.33g

Mass of H_2O=4.366g

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 21.33 g of carbon dioxide, $\frac{12}{44}\times 21.33=5.82g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 4.366 g of water, $\frac{2}{18}\times 4.366=0.485g$ of hydrogen will be contained.

Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles$

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 0.485 moles.

For Carbon = $\frac{0.485}{0.485}=1$

For Hydrogen = $\frac{0.485}{0.485}=1$

For Oxygen = $\frac{0.969}{0.485}=1.99\approx 2$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_1H_{1}O_2=CHO_2$

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 = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

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

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Thus, the molecular formula for the given organic compound is C_2H_2O_4

Answer from: Quest

a. they must be certified by the avma.

explanation:

mark brainliest and have a great day!

Answer from: Quest

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

Another question on Chemistry

Chemistry, 22.06.2019 23:00

What is formed when amino acids form long chains or polymerize

Answers: 1

Answer

Chemistry, 23.06.2019 00:30

What is the percent by mass of magnesium sulfate in mgso4.7h2o

Answers: 3

Answer

Chemistry, 23.06.2019 01:00

Which polymers are most closely related? a. protein and nucleic acids b. cellulose and starch c. nucleic acids and starch d. nucleic acids and cellulose

Answers: 2

Answer

Chemistry, 23.06.2019 01:20

Use the de broglie's wave equation to find the wavelength of an electron moving at 7.3 × 106 m/s. show your work. note: h = plank's constant (6.62607 x 10-34 j s)

Answers: 1

Answer

You know the right answer?

A21.82 gram sample of an organic compound containing c, h and o is analyzed by combustion analysis a...

Questions