This self-discovery of mine occurred some time ago, when I was digging for some Chemistry questions to my leisure. Some questions, I admit, were still out of my league, so I had a pretty hard time finding an answerable question that will secure my 10 points; until I found that question.
“Vitamin C (molar mass = 176 g/mol) contains C, H, and O. 4.00g of vitamin C on combustion gives 6.00g CO₂ and 1.632 H₂0. What is the empirical formula of Vitamin C? What is its molecular formula?”
My first impression was that this Chemistry question is not only easy, but typical also. Most probably the words “empirical formula” and “molecular formula” brainwashed me into thinking that this is just a common problem. But since I am terribly bored and had felt the urge to boost up my points, I proceeded to absorb to question, and it turns out that this isn’t quite as easy as it looks.
A typical empirical/molecular formula question will start like this: A sample is said to contain 9.00% oxygen, 8.75% sulfur… and so on. But this question is quite different; there were no percentages stated, and the masses of the elements present in the compound remain anonymous. At first I thought I’m at my wit’s end, but it turns out, thankfully, I’m not. Armed with my knowledge about moles and forming chemical equations, I started deducing what I can about the problem, with the hopes of gaining clues to solving it.
My first approach, which turns out to be the best one, was to find the mass of oxygen gas that was used, applying the law of conservation of mass. Simple arithmetic is required.
mass O₂ = (mass CO₂ + mass H₂O) – mass Vitamin C
mass O₂ = (6.00 g + 1.632 g) – 4.00 g
mass O₂ = 3.3632 g
Next, I made a dummy equation for reference.
x + O₂ → CO₂ + H₂O
I then proceeded to inspect the masses of the reactant and the products. I realized that if you solved for the number of moles of those substances, well, let’s see.
mol Vitamin C = 4.00 g / 176 g/mol = 0.0227 mol Vitamin C
mol O₂ = 3.3632 g / 32.00 g/mol = 0.105 mol O₂
mol CO₂ = 6.00 g / 44.01 g/mol = 0.136 mol CO₂
mol H₂O = 1.632 g / 18.02 g/mol = 0.0906 mol H₂O
From this gathered data, I can conclude that “0.0227 moles of vitamin C reacted with 0.105 moles of oxygen gas to form 0.136 moles of carbon dioxide and 0.0906 moles of water.” I can now put some numbers in my dummy equation.
0.0227x + 0.105O₂ → 0.136CO₂ + 0.0906H₂O
A conventional chemical equation has whole numbers as coefficients. This is easily settled. I divided all coefficients by the one with the least value; 0.0227.
x + 5O₂ → 6CO₂ + 4H₂O
From this point, finding the molecular formula’s surprisingly easy. I just applied the principles of balancing chemical equations. Inspecting the left hand side, I have 10 O atoms. By inspecting the right hand side, I have 6 C atoms, 16 O atoms and 8 H atoms. This one isn’t really hard. The molecular formula that will balance the equation is:
C₆H₈O₆ + 5O₂ → 6CO₂ + 4H₂O
And so, the molecular formula of vitamin C is C₆H₈O₆, while its empirical formula is C₃H₄O₃. Some would probably say that I could’ve just searched the internet for vitamin C’s molecular and empirical formula, but then again, I wouldn’t have discovered this method if I did it. To test whether this method’s really effective, I searched once again for a question of this type. Fortune has been my friend.
“Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H, and O. Combustion of 19.81 mg terephthalic acid produces 41.98 mg CO₂ and 6.45 mg H₂O. If 0.264 mol of terephthalic acid has a mass of 43.8 g, determine the molecular formula for terephthalic acid.”
My steps are still the same. I solved for the mass of terephthalic acid that was used applying the law of conservation of mass. It gave me:
mass t. acid = (41.98 mg + 6.45 mg) – 19.81 mg
mass t. acid = 28.62 mg
I once again made a dummy equation as a guide.
x + O₂ → CO₂ + H₂O
Afterwards, I proceeded to finding the number of moles of the reactants and products. But first, I had to solve for the molar mass of terephthalic acid, since it wasn’t explicitly stated.
mole = mass / molar mass ⇒ 0.264 mol = 43.8 g / x ⇒ x = 43.8 g / 0.264 mol ⇒ x = 166 g/mol
After finding the number of moles of the substances involved in the reaction, I concluded that:
“0.000119 moles of terephthalic acid reacted with 0.000894 moles of oxygen gas to produce 0.000954 moles of carbon dioxide and 0.000360 moles of water.”
You can check my math if you want. Going back to my dummy equation:
0.000119x + 0.000894O₂ → 0.000954CO₂ + 0.000360H₂O
For the sake of simplicity, I dropped the zeroes and reduced the coefficients to lowest terms by dividing all of them by the coefficient with the least value; 119.
x + 7.5O₂ → 8CO₂ + 3H₂O ⇒ 2x + 15O₂ → 16CO₂ + 6H₂O
By inspection, the equation will be balanced if x is C₈H₆O₄. Hence, the molecular formula of terephthalic acid is C₈H₆O₄. I checked the internet if I got it right, and I actually did.
Of course, someone else have discovered this too, but I assure you, I solved these problems on the spot, and, for some reason, there were absolutely no other “answerers.” I just utilized what I know to solve this problem, and fortunately, I succeeded. I’m still not sure how to call this method though, so for now, I’ll call this “the conjecture.”
“Vitamin C (molar mass = 176 g/mol) contains C, H, and O. 4.00g of vitamin C on combustion gives 6.00g CO₂ and 1.632 H₂0. What is the empirical formula of Vitamin C? What is its molecular formula?”
My first impression was that this Chemistry question is not only easy, but typical also. Most probably the words “empirical formula” and “molecular formula” brainwashed me into thinking that this is just a common problem. But since I am terribly bored and had felt the urge to boost up my points, I proceeded to absorb to question, and it turns out that this isn’t quite as easy as it looks.
A typical empirical/molecular formula question will start like this: A sample is said to contain 9.00% oxygen, 8.75% sulfur… and so on. But this question is quite different; there were no percentages stated, and the masses of the elements present in the compound remain anonymous. At first I thought I’m at my wit’s end, but it turns out, thankfully, I’m not. Armed with my knowledge about moles and forming chemical equations, I started deducing what I can about the problem, with the hopes of gaining clues to solving it.
My first approach, which turns out to be the best one, was to find the mass of oxygen gas that was used, applying the law of conservation of mass. Simple arithmetic is required.
mass O₂ = (mass CO₂ + mass H₂O) – mass Vitamin C
mass O₂ = (6.00 g + 1.632 g) – 4.00 g
mass O₂ = 3.3632 g
Next, I made a dummy equation for reference.
x + O₂ → CO₂ + H₂O
I then proceeded to inspect the masses of the reactant and the products. I realized that if you solved for the number of moles of those substances, well, let’s see.
mol Vitamin C = 4.00 g / 176 g/mol = 0.0227 mol Vitamin C
mol O₂ = 3.3632 g / 32.00 g/mol = 0.105 mol O₂
mol CO₂ = 6.00 g / 44.01 g/mol = 0.136 mol CO₂
mol H₂O = 1.632 g / 18.02 g/mol = 0.0906 mol H₂O
From this gathered data, I can conclude that “0.0227 moles of vitamin C reacted with 0.105 moles of oxygen gas to form 0.136 moles of carbon dioxide and 0.0906 moles of water.” I can now put some numbers in my dummy equation.
0.0227x + 0.105O₂ → 0.136CO₂ + 0.0906H₂O
A conventional chemical equation has whole numbers as coefficients. This is easily settled. I divided all coefficients by the one with the least value; 0.0227.
x + 5O₂ → 6CO₂ + 4H₂O
From this point, finding the molecular formula’s surprisingly easy. I just applied the principles of balancing chemical equations. Inspecting the left hand side, I have 10 O atoms. By inspecting the right hand side, I have 6 C atoms, 16 O atoms and 8 H atoms. This one isn’t really hard. The molecular formula that will balance the equation is:
C₆H₈O₆ + 5O₂ → 6CO₂ + 4H₂O
And so, the molecular formula of vitamin C is C₆H₈O₆, while its empirical formula is C₃H₄O₃. Some would probably say that I could’ve just searched the internet for vitamin C’s molecular and empirical formula, but then again, I wouldn’t have discovered this method if I did it. To test whether this method’s really effective, I searched once again for a question of this type. Fortune has been my friend.
“Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticizers. It contains only C, H, and O. Combustion of 19.81 mg terephthalic acid produces 41.98 mg CO₂ and 6.45 mg H₂O. If 0.264 mol of terephthalic acid has a mass of 43.8 g, determine the molecular formula for terephthalic acid.”
My steps are still the same. I solved for the mass of terephthalic acid that was used applying the law of conservation of mass. It gave me:
mass t. acid = (41.98 mg + 6.45 mg) – 19.81 mg
mass t. acid = 28.62 mg
I once again made a dummy equation as a guide.
x + O₂ → CO₂ + H₂O
Afterwards, I proceeded to finding the number of moles of the reactants and products. But first, I had to solve for the molar mass of terephthalic acid, since it wasn’t explicitly stated.
mole = mass / molar mass ⇒ 0.264 mol = 43.8 g / x ⇒ x = 43.8 g / 0.264 mol ⇒ x = 166 g/mol
After finding the number of moles of the substances involved in the reaction, I concluded that:
“0.000119 moles of terephthalic acid reacted with 0.000894 moles of oxygen gas to produce 0.000954 moles of carbon dioxide and 0.000360 moles of water.”
You can check my math if you want. Going back to my dummy equation:
0.000119x + 0.000894O₂ → 0.000954CO₂ + 0.000360H₂O
For the sake of simplicity, I dropped the zeroes and reduced the coefficients to lowest terms by dividing all of them by the coefficient with the least value; 119.
x + 7.5O₂ → 8CO₂ + 3H₂O ⇒ 2x + 15O₂ → 16CO₂ + 6H₂O
By inspection, the equation will be balanced if x is C₈H₆O₄. Hence, the molecular formula of terephthalic acid is C₈H₆O₄. I checked the internet if I got it right, and I actually did.
Of course, someone else have discovered this too, but I assure you, I solved these problems on the spot, and, for some reason, there were absolutely no other “answerers.” I just utilized what I know to solve this problem, and fortunately, I succeeded. I’m still not sure how to call this method though, so for now, I’ll call this “the conjecture.”
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