Question

A .3459g tablet that contains bicarbonate generates 1.24x10^-3 moles of CO2. what is the percent bicarbonate in the tablet??

Answer 1

can you think of a way to relate the moles of bicarbonate \(HCO_3^-\) to the moles of \(CO_2\)?

Answer 2

its a 1:1 mole ratio right?

Answer 3

indeed.

Answer 4

so do i do (.3459/ 1.24x10^-3)x100 ?

Answer 5

no. convert moles of bicarbonate to mass of bicarbonate.

then (mass of bicarbonate/0.3459)x100

Answer 6

oooh i see. so (61.01/.3459)x100 ??

Answer 7

no mass (in grams) not the molar mass

Answer 8

but since there is only 1 mole of HCO3 then it would be the same?

Answer 9

nope. what you're doing is using the moles of CO2 and relating them to HCO3

Since the number of carbon atoms stays constant, they have they same amount of moles.

find the mass of bicarbonate that corresponds to the number of moles of CO2

Answer 10

Ok so there is 1.24x10^-3 moles of HNO3. S

Answer 11

(So i do 1.24x10^-3 )x 61.01g = .07566

Then .07566/.3459=.21

.21x100= 21.87% ?

Answer 12

yep thats it

Answer 13

ok thanks! i just totally forgot that they gave us the moles of CO2