CHEMISTRY

I'm trying to figure out the percent composition of a mixture of an unknown (a combination of NaHCO3 and NaCl).

Here's my data:
Mass of 150mL of empty beaker: 80.937g
Mass of 150mL beaker w/ unknown: 81.957g
Mass of 150mL beaker w/ unknown after reaction: 89.126g
Mass of bottle of 1.00 M HCl: 172.079g
Mass of bottle of 1.00 M HCl after reaction: 164.621g

My idea of finding the percent composition is involving the calculations of turning CO2 mass into moles of CO2 to moles of NaHCO3 to mass of NaHCO3 to get the percent composition of NaHCO3. I'm not quite sure if I have got my calculations down right.

Calculation of CO2 loss (172.079g + 81.957g) - (164.621g - 89.126g) = .289g CO2

Since CO2 and NaHCO3 has a 1:1 we can do this calculation to get the mass of NaHCO3

.289g CO2 * (1 mol/44g CO2) * (84g NaHCO3/1 mol) = .5517g NaHCO3

Then the percent composition of NaHCO3 would be. (Mass of NaHCO3/ Molar mass of NaHCO3) * 100

(.5517g NaHCO3 / 84.00g NaHCO3) * 100 = .6568%

That means I have 99.34% of NaCl in the mixture if .6568% is NaHCO3. This doesn't seem logical to me.

Can you please provide an explanation on what I did wrong or where I went wrong?

Thank you !

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asked by Liam
  1. [Update]

    I think I know what I did wrong. Instead of: (Mass Sample of NaHCO3 / Molar Mass of NaHCO3) * 100 = some value, it should be (Mass Sample of NaHCO3 / Total mass sample of unknown mixture) * 100 = some value.

    So, (.5517g NaHCO3 / 1.020g of unknown mixture) * 100 = 54.09% of NaHCO3

    Please let me know if I have the right idea. Thank you!

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    posted by Liam

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