H3CA is a vitally important intemediate in the metabolism of carbohydrates. The compound is widely distributed in plant and animal tissues and fluids. H3CA is a tiprotric acid, ie, it requires three moles of NaOH to fully neutralize one mole of H3CA.
a.) If it requires 19.41 of 0.616 M NaOH(aq) to fully neutralize 0.766g of H3CA, what is the molar mass of H3CA?
b.) Given the following % composition, determine the true formula of H3CA: 37.51% C; 4.20% H, 58.29% O.
* I wasn't sure where to begin because there are a lot of values to consider. So, what should I start with first?*
The equation.
H3CA + 3NaOH ==> Na3CA + 3H2O
For part b, take 100 g sample, convert to mols C, mols H, mols O, then find the ratio of atoms to each other with the smallest number being no less than 1.0.
Post your work if you get stuck.
For part A. is that just extra info?
Molar mass is simply mass of H *3 plus the mass of Calcium which ='s 43.104 g/mol H3CA..
For part B... given the percent composition, I did
37.51 g/ 12.0111 g/mol= 3.123 mol C
4.20 g/ 1.0079 g/mol= 4.17 mol H
58.29 g/ 15.999 = 3.643 mol O.
Now to find the true formula or Empirical Formula, I'd have to divide those previous answers by the small mole number which is 3.123.
I guess the empirical formula would be:
C1 H1.34 O1.16 ?
Does that make sense and is this even right?
H3CA is a vitally important intemediate in the metabolism of carbohydrates. The compound is widely distributed in plant and animal tissues and fluids. H3CA is a tiprotric acid, ie, it requires three moles of NaOH to fully neutralize one mole of H3CA.
a.) If it requires 19.41 of 0.616 M NaOH(aq) to fully neutralize 0.766g of H3CA, what is the molar mass of H3CA?
b.) Given the following % composition, determine the true formula of H3CA: 37.51% C; 4.20% H, 58.29% O.
* I wasn't sure where to begin because there are a lot of values to consider. So, what should I start with first?*
To determine the molar mass of H3CA and its true formula, you can start by solving part (a) of the question, which involves stoichiometry and titration.
a.) To find the molar mass of H3CA, we need to calculate the number of moles of H3CA and divide it by the mass in grams. We can start by calculating the number of moles of NaOH used in the neutralization reaction.
Given:
- Volume of NaOH(aq) used = 19.41 mL
- Concentration of NaOH(aq) = 0.616 M
To convert the volume into liters, divide it by 1000:
Volume of NaOH(aq) used = 19.41 mL / 1000 = 0.01941 L
Now, we can calculate the number of moles of NaOH:
Moles of NaOH = concentration × volume
Moles of NaOH = 0.616 M × 0.01941 L = 0.01192956 moles
According to the stoichiometry of the reaction, it takes three moles of NaOH to neutralize one mole of H3CA. Therefore, the number of moles of H3CA is one-third of the number of moles of NaOH used.
Moles of H3CA = 0.01192956 moles / 3 = 0.00397652 moles
Finally, to calculate the molar mass, divide the mass of H3CA by the number of moles:
Molar mass of H3CA = Mass of H3CA / Moles of H3CA
Molar mass of H3CA = 0.766 g / 0.00397652 moles
Now that you have calculated the molar mass of H3CA in part (a), let's move on to part (b).
b.) Given the percent composition of H3CA as:
- 37.51% Carbon (C)
- 4.20% Hydrogen (H)
- 58.29% Oxygen (O)
To determine the true formula of H3CA, you need to convert the percent composition into the empirical formula. The empirical formula represents the simplest whole number ratio of elements in the compound.
Let's assume we have 100 grams of H3CA, which would correspond to the given percent composition. We can then convert these masses into moles.
- Moles of carbon (C) = (37.51 g / 12.01 g/mol)
- Moles of hydrogen (H) = (4.20 g / 1.01 g/mol)
- Moles of oxygen (O) = (58.29 g / 16.00 g/mol)
Next, divide each of the calculated moles by the smallest calculated mole value to obtain a whole number ratio. In this case, the smallest value is the moles of carbon (C).
Dividing by 0.003122, which is the number of moles of C:
- Moles of C = (37.51 g / 12.01 g/mol) / 0.003122 mol = 12.02 ≈ 12
- Moles of H = (4.20 g / 1.01 g/mol) / 0.003122 mol = 1.34 ≈ 1
- Moles of O = (58.29 g / 16.00 g/mol) / 0.003122 mol ≈ 3.72 ≈ 4
By dividing each of the moles by the smallest mole value, we obtained the empirical formula as C12H4O4. However, from the information provided in the question, we know that H3CA requires three moles of NaOH to neutralize one mole of H3CA. Therefore, we need to multiply the empirical formula by 3 to get the true formula.
The true formula of H3CA is therefore C36H12O12.