A white powder, consisting of a simple mixture of tartaricacid (C 4 H 6 O 6 ) an

Question

A white powder, consisting of a simple mixture of tartaricacid (C4H6O6) andcitric acid(C6H8O7) was analysed to determinethe elemental composition. Combustion of a223.8-mg sample produced 282.8 mgof CO2 and 82.1 mg ofH2O. Use atomic masses: C 12.011; H 1.00794; O 15.9994. calculate the % carbon, hydrogen, oxygen, tartaric acid bymass, in the sample answers are % thank you!! calculate the % carbon, hydrogen, oxygen, tartaric acid bymass, in the sample answers are % thank you!! answers are % thank you!! Use atomic masses: C 12.011; H 1.00794; O 15.9994.

Explanation / Answer

The reaction then can be written as follows: C4H6O6 + C6H8O7 + 7O2 ---> 10CO2 + 7H2O If the moles of C4H6O6 = n mmol, and that of C6H8O7 = mmmol, Since the total moles of carbon from both C4H6O6 and C6H8O7 =the moles of carbon of CO2,
we can write (4 x n) + (6 x m) = 282.8 / 44    (eq.1)
(note: FW of CO2 = 44; 4 x n is moles of C in tartaric acid,whereas 6 x m is that in citric acid)
and
since the total moles of hydrogen from both C4H6O6and C6H8O7 = the moles of hydrogen of H2O
Then, (6 x n) + (8 x m) = 2 x (82.1 / 18)   (eq. 2)
(note: FW of H2O = 18; 6 x n = moles of H in tartaric acid, 8x m = that in citric acid)
by Solving equation 1 and 2, we get n = 0.82 mmol, it is themole number of tartaric acid.
Thus,  the % carbon, hydrogen, oxygen, tartaric acidby mass in the sample (223.8 mg) are
The % C = (0.82 mmol x 4 x 12.011mg/mmol) / 223.8 mg = The % H = (0.82 mmol x 6 x 1.00794mg/mmol / 223.8 mg = The % O = (0.82 mmol x 6 x 15.9994mg/mmol) / 223.8 mg =

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.