1. Acetic acid contains the elements carbon, hydrogen, and oxygen. When 5.00 g o

Question

1. Acetic acid contains the elements carbon, hydrogen, and oxygen. When 5.00 g of acetic acid are burned in air, 7.33g of CO2, and 3.00g of water are obtained. Determine the empirical formula of acetic acid. (5 points)

2.

For the equation,                Al    +      O2           ----->                         Al2O3             answer the following questions: (5 points)

a. Balance the above equation

b. If 10.0 g of Al are used, how many g of oxygen are needed and how many g of product are produced?

c. If you had 20 moles of Al, how many moles of oxygen would be needed and how many moles of product would be produced?

d. Based on the information in b., if you isolated 5.0 g of Al2O3, calculate the % yield.

                 e. If you had 5.0 g of Al and 4.5 g of O2, which is the limiting reagent?

Explanation / Answer

Q1 Solution

Given data

Mass of acetic acid burned = 5.00 g acetic acid

Mass of CO2 formed = 7.33 g

Mass of H2O formed = 3.00 g

Empirical formula of acetic acid = ?

From the given masses of the CO2 and H2O first calculate the moles of the CO2 and H2O and then using the moles of the C and CO2 and mole ratio of the H and H2O we can calculate the moles of the C and H as follows

Moles = mass / molar mass

Moles of CO2 = 7.33 g / 44.01 g per mol   = 0.1666 mol CO2

Since mole ratio of the CO2 to C is 1 :1 so moles of C = 0.1666 mol

Moles of H2O = 3.00 g / 18.0148 g per mol = 0.1666 mol H2O

Since mole ratio of the H2O to H is 1 :2 therefore moles of H = 0.1666 * 2 = 0.3332 mol H

From the moles of the C and H we need to calculate the mass of the C and H and then subtract it from the mass of the acetic acid to get the mass of oxygen present in the 5.00 g acetic acid sample.

lets calculate mass of C and H

formula to calculate mass is as follows

Mass = moles * molar mass

Mass of C = 0.1666 mol * 12.01 g per mol = 2.00 g

Mass of H =0.3332 mol * 1.0079 g per mol = 0.336 g H

Now lets calculate mass of O in the 5.00 g acetic acid sample

Mass of oxygen = acetic acid mass – (mass of Carbon + mass of hydrogen )

Mass of oxygen = 5.00 g – (2.00 g + 0.336 g)

                             = 2.664 g O

Now using the calculated mass of the oxygen calculate moles of the Oxygen

Moles of Oxygen = 2.664 g / 16.0 g per mol

                                 = 0.1665 mol O

Now using the moles of the each elements we need to calculate the mole ratio of the elements in the sample as follows,

Divide moles of each element by the smallest mole value, then we get

C= 0.1666 / 0.1665 = 1

H= 0.3336 / 0.1665 = 2

O = 0.1665 / 0.1665 = 1

Therefore the ratio of the moles of the elements in sample is 1:2:1 that is 1C ,2H and 1O

Therefore the empirical formula of the acetic acid = CH2O

Q2 ) a) While writing the balanced reaction equation we use the proper coefficient to the reactant and products so that all elements get balanced on the both sides of the reaction equation.

balanced reaction equation is as follows

4Al +3O2 ----> 2Al2O3

b) Lets calculate the mass of oxygen needed to react with 10.0 g Al and calculate the amount of Al2O3 produced.

Using the mole ratio of the reactant and product from the balanced reaction we can calculate the mass of the O2 and Al2O3 as follows Lets calculate mass of O2 needed

Mole ratio of the Al to O2 is 4 :3

Using this mole ratio lets calculate mass of O2 needed

(10.0 g Al * 1 mol / 26.98 g ) * (3 mol O2/4 mol Al) *(32.0 g / 1 mol O2) = 8.89 g O2

Lets calculate mass of the Al2O3 produced

Using this mole ratio lets calculate mass of Al2O3 produced as follows,

(10.0 g Al * 1 mol / 26.98 g )*(2 mol Al2O3 / 4 mol Al ) * (101.96 g/ 1 mol Al2O3) = 18.9 g Al2O3

c) 20 mol Al

mole of O2 = ?

mole of Al2O3 = ?

Here also need to use the mole ratio of the Al with O2 and Al2O3 to calculate moles of the O2 and Al2O3

Mole ratio of the Al to O2 is 4:3

lets calculate moles of O2            

20 mol Al * 3 mol O2 / 4 mol Al = 15 mol O2

Mole ratio of the Al to Al2O3 is 4 :2

Lets calculate moles of Al2O3

20 mol Al * 2 mol Al2O3 / 4 mol Al = 10 mol Al2O3

d) Given data

Actual yield = 5.00 g

theoretical yield from part b = 18.9 g

formula to calculate the percent yield is as follows

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

Lets put the values in the formula and calculate the percent yield

% yield = (5.00 g / 18.9 g)*100%

             = 26.5 %

e) given data

mass of Al = 5.0 g Al

mass of O2 = 4.5 g O2

limiting reactant = ?

Lets first calculate moles of the each reactant using their given mass and molar mass

moles of Al = (5.0 g / 26.98 g.mol-1)

                   = 0.1853

moles of O2 = (4.5 g / 32 g.mol-1) = 0.141 mol O2

Now using the mole ratio of the O2 and Al lets calculate moles of Al needed to react with 0.141 mol O2

(0.141 mol O2 * 4 mol Al) / 3 mol O2 = 0.188 mol Al

Moles of Al needed are 0.188 mol and moles of Al we have are 0.1853 mol

Therefore moles of Al are less than actually needed for the reaction with O2 therefore Al is the limiting reactant.

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