A white powder, consisting of a simple mixture of tartaric acid (C4H6O6) and cit

Question

A white powder, consisting of a simple mixture of tartaric acid (C4H6O6) and citric acid (C6H8O7) was analysed to determine the elemental composition. Combustion of a 364.0-mg sample produced 479.9 mg of CO2 and 135.0 mg of H2O. Use atomic masses: C 12.011; H 1.00794; O 15.9994.

1. Calculate the % carbon, by mass, in the sample.

2. Calculate the % hydrogen, by mass, in the sample.

3. Calculate the % oxygen, by mass, in the sample.

4. Calculate the % citric acid, by mass, in the sample

Assign formal oxidation numbers (also known as oxidation states) to the specified atoms in the following compounds, according to convention.

The N atom in NO2
The P atom in PF5

The Li atom in Li3N

The N atom in NH3

Explanation / Answer

Step 1

Convert all given masses into g.

Mass of sample = 0.364 g

m CO2 = 0.4799 g

m H2O = 0.135 g

Step 2

Calculate molar masses

Molar mass of H2O = 18.0148 g/mol

CO2 : 44.0098 g/mol

Tartaric acid = 150.1 g/mol

Citric acid = 192.1 g/mol

Step 3

Calculate moles of H2O and CO2

Mol H2O = 0.135 g / 18.0148 g per mol = 0.007494 mol

Mol CO2 = 0.4799 g / 44.0098 g per mol = 0.010904 mol

Step 3

Calculate mass and mass percent of C , H and O

Mass of C = moles of CO2 x 1 mol C/ 1 mol CO2 x 12.011 g/mol

= 0.0109 mol CO2 x ( 1 mol C / 1 mol CO2 ) x 12.01 g/mol

= 0.131 g C

Mass percent of C = (0.131 g/ 0.364 ) x 100 = 36.0 %

Mass of H = 0.007494 mol H2O x 2 mol H / 1 mol H2O x 1.00794 g/mol

= 0.015107 g H

Mass percent of H = 0.015107 g/ 0.364 g x 100 = 4.1 %

Mass of O = 0.364 – mass of H – mass of C

= 0.364 g – 0.015107 g H – 0.131 g C

= 0.218 g

Mass percent of O = (0.218/0.364) x100 = 59.9 %

Step 4

Calculation of mass percent of Citric acid.

Lets

x ( mass of tartartic acid) + y ( mass of citric acid ) = 0.364 g     ….. (I)

we also know that there are 4 moles of C in tartaric acid and 6 in citric acid so,

4 * (x/molar mass of tartaric acid) + 6 x ( y / molar mass of citric acid) = 0.010904 ( moles of C ) …(II)

Lets plug in the value of molar masse of tartaric and citric acid

4 * ( x/ 150.1 ) + 6 * ( y / 192.1) = 0.010904 ….(III)

Lets rewrite equation number ( I ) in terms of mass of tartaric acid

Mass of tartaric acid ( x) = 0.364 – mass of citric acid ( y) …..( IV)

Lets put equation (IV) in equation III

(4 ( 0.364 – y) / 150.1 ) + (6 * (y / 192.1)) = 0.010904      …(V)

Lets solve for mass of citric acid ( y)

y = 0.263 g

mass of citric acid = y = 0.263 g

mass percent of citric acid = (0.263/0.364) x 100 = 72.3 % Citric acid

Part II

Oxidation state of O is -2 therefore

O.S of N in NO2

O.S of N + 2 x ( -2) = 0

O.S of N = +4

Oxidation state of F is -1

O.S of P + 5 x (-1) = 0

O.S of P = +5

Li is alkali metal and therefore it has always +1 oxidation state.

N in NH3

Oxidation state of H is +1

Therefore,

Oxidation state of N + 3 x (+1) = 0

Oxidation state of N = -3

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