In the combustion analysis of 0.1127 g of glucose (c6h12o6), what mass, in grams, of co2 would be produced?

Combustion analysis is the basic analytical method of natural solids and fluids. The quantity of carbon-hydrogen, oxygen, and even nitrogen oxides in the substances could be determined. Joseph Louis Gay-Lussac had also invented that method, and the calculation can be defined as follows:

According to the reaction:

of glucose on combustion gives of is partial

of the glucose on combustion  given ......... grams of

Therefore the final mass is "0.1652 g".

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The combustion produces 0.1652 g CO₂.

Balanced equation = C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

Moles of C₆H₁₂O₆ = 0.1127 g C₆H₁₂O₆ × (1 mol C₆H₁₂O₆/180.16 g C₆H₁₂O₆)

= 0.000 6256 mol C₆H₁₂O₆

Moles of CO₂ = 0.000 6256 mol C₆H₁₂O₆× (6 mol CO₂/1 mol C₆H₁₂O₆)

= 0.003 753 mol CO₂

Mass of CO₂ = 0.003 753 mol CO₂ × (44.01g CO₂/1 mol CO₂) = 0.1652 g CO₂

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