Pewien kwas karboksylowy zawiera 64,62% węgla. Oblicz jego masę.
Proszę o szybką odpowiedź
m CnH2n+1COOH = 14n+46
m C = 12n+12
14n+46-------100%
12n+12--------64,62%
64,62(14n+46) = 100(12n+12)
904,68n+2972,52 = 1200n+1200
295,32n = 1772,52
n = 6
C6H13COOH (kwas heptanowy)
mC6H13COOH=130u
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m CnH2n+1COOH = 14n+46
m C = 12n+12
14n+46-------100%
12n+12--------64,62%
64,62(14n+46) = 100(12n+12)
904,68n+2972,52 = 1200n+1200
295,32n = 1772,52
n = 6
C6H13COOH (kwas heptanowy)
mC6H13COOH=130u