Question

2) Refer to the following information for parts a-c: Ascorbic acid, commonly known as Vitamin C, is an essential nutrient involved in tw o and the production of neurotransmitters. Diets deficient in Vitamin C can lead to a condition called survy Combustion analysis of a 12.50 g sample of ascorbic acid produces 18.74 g of Co, and 5.11 H.O. The molecular mass of ascorbic acid is 1 76.12 g/ mol a. Determine the empirical formula for ascorbic acid: b. Determine the molecular formula for ascorbic acid: c. Write and balance the chemical equation for the combustion of ascorbic acid.

Answer 1

Are Mass of copa 18.749 Molar mass of cos= 44.01g/mol Moles of Co2 = Mass of ca Molar mass ogcas = 18.748 44.01 sl mol = 0.425812 315 mol Mass of H₂O = 5.11g Molae mass og 120- 18. 015289/mol

Moles of H₂o- Mass of H2O Molar mass of Ho = 5.11g. 18.01528 g/mol = 0.2836 48103mol We know that I mol CO₂ contains Imola Thus, Moles og C= Moles of con = 0.425812315 mol Molae mass of c= 12.0107 g/mol Mass of C= Moles gcx Molae massage = 0.425812 315 ml x 12.01079 mor = 5.1143 03978 Imol H2O contains a mold. Thug, Molug a = 2x Moleg a 1,0 = (2x 0.283648103)mo = 0.567296206 mol - Mass of H= Moles gux clar mass off = 0.567296 206 md x 1.00784 g/mol

= 0.57174 3808 g Mags og ascorbic acid = 12.509 Mass g o= mass of ascoobic acid -(Mags og ut mass og c)

= 12.50g -(0.5717438089 + 5.114 303979) - 6-81 3952229 Moles gou Mass go Molar mass go = 6.813952228 19.999 g/mol (Melar mass go= 15.9999/mol) = 0.425898632 mol determine simplest whole number ratio. -To

Since Moles of C is the lowest among the moles of H, cando, we devide moles of hand o by moles of C. C: 0.425812315 0.425812 315 = 1.00 M: 0.567296206 0.425812315 in 3202.682 0 : 0 4258986321.000 2027|| 0.425812315 Mattiplying the relative amounts of cl and oby 3 we have, c = (1.00x3)=3 M = (1.33x3) = 4 0 - (1.00x3) = 3 (since we cannot have no. of atoms as fraction in a compound)

- Empirical formula = Cg Ka Oz. empiaical formula mass = (3x 12.0107 + 4x1.00784 + 3x 15.999) 9|mol ( 3x atompe mass og c + 4x a tompe mass ght 3x atompe mass go) - 88.06046 g/mol given: moleculas mass Cascosbpe acid) = 176.12 9/mol Thug, empidical formula units = Molecular mags empia cal formula maes = 176.12 g/mol 88-86046 g/mol 2 Thus, Molecular for mula = empirical formula units x empirical formula - 2x (C34403) = c.no

© Cam bustion of 486: (648O6 +0 Co2 TH2O - A) for balancing the above equation, we add coefficient 5 before Og and coefficient 6 before coz and Coefficient & before kan Thus ean. (A) becomesi GM8 O6 +502 602 + 4H2O.