In a laboratory, 1.55mg of an organic compound containing carbon, hydrogen, and oxygen is burned for analysis. this combustion resulted in the formation of 1.45mg of carbon dioxide and .89 mg of water. what is the empirical formula for this compound?

CH₃O  

Step-by-step explanation:

1. Calculate the mass of each element

Mass of C =  1.45 mg CO₂ × (12.01 mg C/44.01 mg CO₂)   = 0.3957 mg C

Mass of H = 0.89 mg H₂O × (2.016 mg H/18.02 mg H₂O) = 0.0996 mg H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.55 – 0.3957 – 0.0996) mg = 1.055 mg

2. Calculate the moles of each element

Moles of C = 0.3957 mg C × 1mmol C/12.01 mg C   = 0.03295 mol C

Moles of H = 0.0996 mg H × 1 mmol H/1.008 mg H = 0.0988   mol H

Moles of O = 1.055 mg O   × 1 mmol O/ 16.00 mg O = 0.06592 mol O

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

C:  0.032 95/0.032 95 = 1

H:     0.0988/0.032 95 = 2.998

O: 0.065 92/0.032 95  = 2.001

4. Round the ratios to the nearest integer

C:H:O = 1:3:2

5. Write the empirical formula

The empirical formula is CH₃O.

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation:

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