ulate the percentage by mass of the indicated elemo s used in welding; (b) hydro

Question

ulate the percentage by mass of the indicated elemo s used in welding; (b) hydrogen in ascorbic acld, HCH. o a substance used as a nitrogen fertilizer, toxin, C41H64013 compounds: (a) carbon in acetylene, C2H2, a C: (c) hydrogen in ammonium sul. also known as vitamin (NH4)2S04, a fate, (N in PtCl2(NHs)a, a chemotherapy agent called di platinum cäsnlatin; (e) oxygen in the female sex hormone estradiol GaHz02i f) carbon in capsaicin, CsH2/NOs, the com- und that gives the hot taste to chili peppers. 3.27 Bésed on the following structural formulas, calculate the pexcentage of carbon by mass present in each compound: Benzaldehyde -(alond fragrance) (a) H-Q C-C HyCO Vanillin (vanilla flavor) (b) HO-CC -C-H C-C Isopentyl acetate (banana flavor) 3.28 Calculate the percentagE compound of carhon hu

Explanation / Answer

3.26

(a) C in C2H2

C = 12
H = 1

Mass of C in C2H2 = 2 x 12 = 24
Total mass of C2H2 = (2 x 12) + (2 x 1) = 26

So, % of C in C2H2 = [ (Mass of C in C2H2) / (Total mass of C2H2) ] x 100
= (24 / 26) x 100
= 92.31 %

(b) H in HC6H4O6 ( or C6H5O6)

C = 12
H = 1
O = 16

Mass of H in C6H5O6 = 5 x 1 = 5
Total mass of C6H5O6 = (6 x 12) + (5 x 1) + (6 x 16) = 173

So, % of H in C6H5O6 = [ (Mass of H in C6H5O6) / (Total mass of C6H5O6) ] x 100
= (5 / 173) x 100
= 2.89 %

(c) H in (NH4)2SO4 (or N2H8SO4)

N = 14
H = 1
S = 32
O = 16

Mass of H in N2H8SO4 = 8 x 1 = 8
Total mass of N2H8SO4 = (2 x 14) + (8 x 1) + (1 x 32) + (4 x 16) = 132

So, % of H in N2H8SO4 = [ (Mass of H in N2H8SO4) / (Total mass of N2H8SO4) ] x 100
= (5 / 132) x 100
= 3.79 %

(d) Pt in PtCl2(NH3)2 (or PtCl2N2H6)

Pt = 195
Cl = 35.5
N = 14
H = 1

Mass of Pt in PtCl2N2H6 = 195 x 1 = 195
Total mass of PtCl2N2H6 = (1 x 195) + (2 x 35.5) + (2 x 14) + (6 x 1) = 300

So, % of Pt in PtCl2N2H6 = [ (Mass of Pt in PtCl2N2H6) / (Total mass of PtCl2N2H6) ] x 100
= (195 / 300) x 100
= 65 %

(e) O in C18H24O2

C = 14
H = 1
O = 16

Mass of O in C18H24O2 = 2 x 16 = 32
Total mass of C18H24O2 = (18 x 12) + (24 x 1) + (2 x 16) = 272

So, % of O in C18H24O2 = [ (Mass of O in C18H24O2) / (Total mass of C18H24O2) ] x 100
= (32 / 272) x 100
= 11.76 %

(f) C in C18H27NO3

C = 14
H = 1
N = 14
O = 16

Mass of C in C18H27NO3 = 18 x 12 = 216
Total mass of C18H27NO3 = (18 x 12) + (27 x 1) + (1 x 14) + (3 x 16) = 305

So, % of C in C18H27NO3 = [ (Mass of C in C18H27NO3) / (Total mass of C18H27NO3) ] x 100
= (216 / 305) x 100
= 70.82 %

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