Answer to Question #164364 in General Chemistry for Lauren Barad

Question

A student found an unknown white powder in his refrigerator. His mom tells him that it is either baking soda or baking powder. He decided to try to figure out what it is through analytical chemistry.

1. Baking soda is simply sodium bicarbonate (NaHCO₃), so to begin the student set out to compare the two by percent by mass. What is the percent Na, H, C, and O in sodium bicarbonate?

a.    %_Na

b.    %_H

c.    %_c

d.    %_O

2.The student then looked up the difference between the two on the internet. It seems baking powder is a mixture of multiple substances (NaHCO3, KHCO₃, KC₄H₅O₆, etc), while baking soda is a pure compound. Based on this would baking powder have a higher, lower, or similar percent sodium by mass? Explain.

3.The student brings 5.00g of the powder to his chemistry teacher who takes the powder to his stock room for a few minutes and returns to tell him that he powder contains 3.58 x 10²² atoms of C – the chemistry teacher is not very helpful…

How many grams of C is that? 

4.What percent of the 5.00g of white powder is C?

5.The student predicts that the white powder IS baking soda (NaHCO₃ – a pure substance). Do you agree with the student, or disagree and why?

Answer 1

1.

NaHCO₃ = 84g/mol

%Na = 27.38%

%H = 1.19%

%C = 14.28%

%O = %57.14%

2 .

The amount of Na in the baking powder will be less than the that of baking soda. Potassium ion will have a potion in the compound.

3.

Mole = No of atoms/avogradro constant

= 3.58 x 10²²/6.023 x 10²³

= 0.0594mol

mass = mole x molar mass

= 0.0594mol x 12.0g

= 0.7128g

4.

%c = 0.7128g/5.00g

= 14.2%

5.

I do because this is no percentage of impurity. The %C of the pure one above is approximately equal to what we have.