Determine the percent yield of water if 2.79 g of water is actually produced fro

Question

Determine the percent yield of water if 2.79 g of water is actually produced from the combustion of 5.35 g of C6H12O6 with 6.19 g of oxygen.

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A compound containing only C, H and O was subjected to combustion analysis. A sample of 4.120 g produced 7.444 g of CO2 and 2.539 g of H2O. Determine the empirical formula of the compound and enter the appropriate subscript after each element.

C?H?O?

If the molar mass of the compound is 73.071 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element.
C?H?O?

Explanation / Answer

Determine the percent yield of water if 2.79 g of water is actually produced from the combustion of 5.35 g of C6H12O6 with 6.19 g of oxygen.

Solution :-

Balanced reaction equation

C6H12O6 + 6O2 ------ > 6CO2 + 6H2O

Now lets calculate the mass of the H2Oproduced from each reactant using the mole ratio

Calculating from the C6H12O6

(5.35 g C6H12O6 * 1 mol / 180.15 g)*(6 mol H2O/ 1 mol C6H12O6)*(18.015 g / 1 mol H2)=

3.21 g water

Now lets calculate from the 6.19 g O2

(6.19 g O2 * 1 mol / 32 g)*(6 mol H2O / 6 mol O2)*(18.015 g / 1 mol H2O)= 3.49 g H2O

C6H12O6 gives the less mass of the H2O therefore theoretical yield of the H2O = 3.21 g

Now lets calculate the percent yield

% yield = (actual yield / theoretical yield )*100%

               = (2.79 g/3.21 g)*100%

               = 86.92 %

A compound containing only C, H and O was subjected to combustion analysis. A sample of 4.120 g produced 7.444 g of CO2 and 2.539 g of H2O. Determine the empirical formula of the compound and enter the appropriate subscript after each element.

Solution :-

Lets first calculate the moles of the CO2 and H2O

Moles = mass / molar mass

Moles of CO2 = 7.444 g / 44.01 g per mol = 0.1691 mol CO2

So the moles of the C = 0.1691 mol since mole ratio is 1 : 1

Moles of H2O = 2.539 g / 18.015 g per mol = 0.141 mol H2O

Moles of H = 0.141 mol H2O * 2 mol H / 1 mol H2O = 0.282 mol H

Now lets calculate the mass of the C and H

Mass of C= 0.1691 mol * 12.01 g per mol = 2.03g

Mass of H = 0.282 mol * 1.0079 g per mol = 0.2482 g

Now lets find mass of oxygen

Mass of oxygen = 4.120 g – ( mass of C + mass of H)

                            = 4.120 g – (2.03 g +0.2482 g)

                          = 1.842 g

Now lets calculate the moles of the O

Moles of O = 1.842 g / 16 g per mol = 0.115 mol O

Now lets find the mole ratio of the each element

C= 0.1691 mol / 0.115 mol = 1.5

H= 0.282/0.115 = 2.5

O = 0.115 / 0.115 = 1

To get the integer number lets multiply the ratios by 2

C= 1.5*2 = 3

H= 2.5*2 = 5

O=1*2 = 2

So the empirical formula is C3H5O2

If the molar mass of the compound is 73.071 g/mol, determine the molecular formula of the compound and enter the appropriate subscript after each element.

Solution :-

Lets find the ratio of the molecular mass to the empirical formula mass

Empirical formula mass (C3H5O2) = 73.07 g per mol

Molar mass / empirical formula mass = 73.07 g per mol / 73.07 g per mol = 1

Therefore the molecular formula = 1 * empirical formula

                                                          = 1*C3H5O2

                                                          = C3H5O2

Both empirical formula and molecular formula are same for this compound.

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