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.
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?
For part a. Is WHAT just extra info?
For part B. Two points.
1. H3CA is just a name given for the triprotic acid in the problem. The CA is NOT (repeat, NOT) calcium.
2. You're on the right track for the empirical formula but you forgot that the subscripts must be whole numbers. When you get strange looking numbers like you have (which by the way are correct), you try to round them to whole numbers BUT you may not round more than maybe 0.1 or so. Since these are out of that range, here is what you do. Multiply by whole numbers to see if you can get another set of whole numbers. Like this.
C1
H1.34
O1.16 Multiply by 2 to get
C2H2.68O2.32. Nope, won't do it so try multiplying by 3 to get this.
C3H4O3.5 and when I spot that I quickly note that the 3.5 multiplied by 2 gives me 7 (a whole number) so I multiply everything by another 2 to get
C6H8O7 and that is the empirical formula. I expect that is citric acid.
For part A, the given information is necessary to calculate the molar mass of H3CA. You correctly mentioned that H3CA requires three moles of NaOH to fully neutralize one mole of H3CA.
To find the molar mass of H3CA, you can use the equation:
Molar mass (H3CA) = (mass of NaOH) / (volume of NaOH)
Given the mass of H3CA (0.766 g) and the concentration of NaOH (0.616 M), you can calculate the volume of NaOH using the formula:
Volume of NaOH = (mass of NaOH) / (molar mass of NaOH) / (concentration of NaOH)
Substituting the given values, the volume of NaOH is:
Volume of NaOH = 19.41 mL
Since three moles of NaOH are required to neutralize one mole of H3CA, the moles of H3CA can be calculated as:
Moles of H3CA = (volume of NaOH) / 3
Moles of H3CA = 19.41 mL / 3 = 6.47 mmol
Now, you can calculate the molar mass of H3CA using the formula:
Molar mass (H3CA) = (mass of H3CA) / (moles of H3CA)
Substituting the given values:
Molar mass (H3CA) = 0.766 g / 6.47 mmol
Molar mass (H3CA) = 118.5 g/mol
So, the molar mass of H3CA is 118.5 g/mol.
For part B, you correctly calculated the moles of each element (C, H, and O) using the percent composition.
Dividing each mole value by the smallest mole value, which is 3.123 (the moles of C), you get:
C: 3.123 / 3.123 = 1
H: 4.17 / 3.123 ≈ 1.34
O: 3.643 / 3.123 ≈ 1.17
The empirical formula based on these ratios is C1H1.34O1.17. Since subscripts in a formula should be whole numbers, you can multiply all the subscripts by 3 to get whole numbers:
C3H4O3
So, the true formula or empirical formula of H3CA is C3H4O3.