Table salt, NaCl, and sugar C12H22O11 are accidentally mixed. A 5.50g sample is burned and 3.60g of CO2 is produced. What was the mass percentage of the table salt mixture?

Answer 1

M_C12H22O11=342
M_{CO2=44
C12H22O11+xO2=12CO2+...
Ratio of mass of C12H22O11 to CO2
=342:12*44
=342:484
Mass of C12H22O11 that produces 3.6g CO2
=3.6*(342/484)
=2.544g
Mass ratio NaCl:C12H22O11
=(5.5-2.544):2.544
=2.956:2.544
Calculate mass percentage and reduce to 3 significant figures.}

Answer 2

3.60 x (342/12*44) = 3.6 x (342/528) = 2.33 g C12H22O11 to produce 3.60 g CO2.

Then 5.50-2.33 = 3.17 g NaCl and
%NaCl = (3.17/5.50)*100 = ?

Answer 3

57.6 %

Answer 4

3.60 x (342/12*44) = 3.6 x (342/528) = 2.33 g C12H22O11 to produce 3.60 g CO2.

Then 5.50-2.33 = 3.17 g NaCl and
%NaCl = (3.12/5.59)*100 = ?

Answer 5

70.44