The moles of each element in the organic compound can be calculated from the mass of the products.
This is the method we use to find the empirical formula of an organic compound, and here is the equation:
CxHy + zO2 + xCO2 + y/2 H2O
This is the balanced chemical equation for the burning of CxHy.
The moles of C and H in CxHy = moles of C in CO2 and the moles of H in H2O.
Now let's apply this into an example problem.
Ex/ When 3.79 grams of an organic compound is burned, 6.61 grams of CO2 and 3.59 grams of H2O is produced. What is the empirical formula of this compound?
Step 1: Find moles of CO2 and H2O
6.61 grams CO2 x 1 mol CO2 / 44 grams CO2 = 0.150 mol CO2
3.59 grams H2O x 1 mole H2O / 18 grams H2O = 0.199 mol H2O
Step 2: Find moles of C and H
0.150 x 1mol C/1 mol CO2 = 0.150 mol C
There is one C in CO2, so we multiply by 1 getting the same result but changing the unit.
0.199 x 2 mol H/1 mol H2O = 0.399 mol H
As for H, there are 2H in H2O so we simply multiply by 2.
Step 3: Check masses
If the mass of C and H does not equal the mass of the compound, then there must be a component of Oxygen present in the compound.
Therefore, mass of Oxygen = mass of compound - mass of C and H
0.150 mol C x 12 grams C / 1 mol C = 1.80 grams C
0.399 mol H x 1 grams H / 1 mol H = 0.399 grams H
1.80 grams + 0.399 grams = 2.199 grams
Now we need to find out the mass of Oxygen.
3.79 grams - 2.199 grams = 1.591 grams O
1.591 grams O x 1 mol O / 16 grams O = 0.099 mol O
Step 4: Find ratio + change ratio into whole number
Just like what we have learned in the previous lesson, divide each molar amount by the smallest molar amount.
C: 0.150 mol / 0.099 mol = 1.5
H: 0.399 mol / 0.099 mol = 4
O: 0.099 mol / 0.099 mol = 1
Now we need to make these into whole numbers.
C: 1.5 x 2 = 3
H: 4 x 2 = 8
O: 1 x 2 = 2
Final answer: C3H8O2
Step 1: Find moles of CO2 and H2O
6.61 grams CO2 x 1 mol CO2 / 44 grams CO2 = 0.150 mol CO2
3.59 grams H2O x 1 mole H2O / 18 grams H2O = 0.199 mol H2O
Step 2: Find moles of C and H
0.150 x 1mol C/1 mol CO2 = 0.150 mol C
There is one C in CO2, so we multiply by 1 getting the same result but changing the unit.
0.199 x 2 mol H/1 mol H2O = 0.399 mol H
As for H, there are 2H in H2O so we simply multiply by 2.
Step 3: Check masses
If the mass of C and H does not equal the mass of the compound, then there must be a component of Oxygen present in the compound.
Therefore, mass of Oxygen = mass of compound - mass of C and H
0.150 mol C x 12 grams C / 1 mol C = 1.80 grams C
0.399 mol H x 1 grams H / 1 mol H = 0.399 grams H
1.80 grams + 0.399 grams = 2.199 grams
Now we need to find out the mass of Oxygen.
3.79 grams - 2.199 grams = 1.591 grams O
1.591 grams O x 1 mol O / 16 grams O = 0.099 mol O
Step 4: Find ratio + change ratio into whole number
Just like what we have learned in the previous lesson, divide each molar amount by the smallest molar amount.
C: 0.150 mol / 0.099 mol = 1.5
H: 0.399 mol / 0.099 mol = 4
O: 0.099 mol / 0.099 mol = 1
Now we need to make these into whole numbers.
C: 1.5 x 2 = 3
H: 4 x 2 = 8
O: 1 x 2 = 2
Final answer: C3H8O2
The above is a pretty good video to review the last lesson as well as what we learned in this lesson.
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