Citric acid, the compound responsible for the sour taste of lemons, has the following elemental composition: C{\rm C}, 37.51%\%; H{\rm H}, 4.20%\%; O{\rm O}, 58.29%\%.Calculate the empirical formula of citric acid
Take a 100 g sample. This gives you
37.51g C.
58.29g O
4.20g H.
Convert to mols.
mols = grams/molar mass
37.51/12 = approx 3.13
58.29/16 = approx 3.64
4.20/1 = approx 4.20
Now find the ratio of the elements to each other with the smallest whole number being 1.00. The easy way to do that is to divide all of the numbers by the smallest value; in this case that is 3.13.
C = 3.13/3.13 = 1.000
O = 3.64/3.13 = 1.16
H = 4.20/3.13 = 1.34
You may not round these numbers to whole numbers. What you do is multiply each by a whole number, startinhg with 2 and progressing by integers until you get a set of whole number.
by 2 gives 2:2.32:2.68
by 3 gives 3:3.48:4 and that tells you that by multiplying by 6 that 3.5 will become a whole number so
by 6 gives 6:7:8 so the empirical formula is C6H8O7
Oh, citric acid, always sour and keeping lemons on their toes! Let me grab my test tubes and lab coat, and we'll get down to some chemistry fun.
To find the empirical formula of citric acid, we first need to find the ratio of atoms present in the compound. We do this by assuming we have 100 grams of citric acid (gotta love those round numbers!).
Given that carbon (C) makes up 37.51% of the compound, we have 37.51 grams of carbon. Similarly, with hydrogen (H) at 4.20% and oxygen (O) at 58.29%, we have 4.20 grams of hydrogen and 58.29 grams of oxygen.
Now, let's find the number of moles for each element. We divide the mass in grams by the molar mass (atomic weight) of the element.
For carbon, the molar mass is around 12 g/mol, so we have 37.51 g / 12 g/mol = 3.125 mol.
For hydrogen, the molar mass is about 1 g/mol, so we have 4.20 g / 1 g/mol = 4.20 mol.
As for oxygen, with a molar mass of approximately 16 g/mol, we have 58.29 g / 16 g/mol = 3.643 mol.
To simplify the ratio of these atoms, we divide each number of moles by the lowest number of moles among them, which is 3.125.
For carbon, we get 3.125 mol / 3.125 mol = 1.
For hydrogen, we get 4.20 mol / 3.125 mol ≈ 1.344.
And for oxygen, we get 3.643 mol / 3.125 mol ≈ 1.165.
Now, we round these numbers to the nearest whole numbers to find the subscripts in the empirical formula.
So, the empirical formula of citric acid is CH1.344O1.165. But since we want whole numbers, we can multiply everything by 4 to get rid of those decimals.
That gives us the empirical formula of citric acid as C4H5O4.
Now, that should clear up the mystery of citric acid's elemental composition. Just remember, when life gives you citric acid, make some lemonade with a twist!
To calculate the empirical formula of citric acid, we need to determine the simplest, whole-number ratio of the elements present in the compound.
1. Start by assuming we have 100 grams of citric acid. This assumption allows us to work with percentages as grams.
2. Convert the percentage composition of each element to grams:
- Carbon (C): 37.51% of 100g = 37.51g
- Hydrogen (H): 4.20% of 100g = 4.20g
- Oxygen (O): 58.29% of 100g = 58.29g
3. Calculate the number of moles of each element using their molar masses:
- Carbon (C): 37.51g / 12.01 g/mol = 3.12 mol
- Hydrogen (H): 4.20g / 1.01 g/mol = 4.15 mol
- Oxygen (O): 58.29g / 16.00 g/mol = 3.64 mol
4. Divide the number of moles of each element by the smallest number of moles to obtain the simplest, whole-number ratio:
- Carbon: 3.12 mol / 3.12 mol = 1
- Hydrogen: 4.15 mol / 3.12 mol = 1.33
- Oxygen: 3.64 mol / 3.12 mol = 1.17
5. Round the ratio to the nearest whole number:
- Carbon: 1
- Hydrogen: 1
- Oxygen: 1
6. Therefore, the empirical formula of citric acid is CH₃O.