The substance that burns in a combustion reaction is called the
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I can't actually answer this one if the empirical formula is not given. Luckily, I've found a similar problem from another website. The problem is shown in the picture attached. It shows that the empirical formula is CH₂O. Let's calculate the molar mass of the empirical formula.
Molar mass of E.F = 12 + 2(1) + 16 = 30 g/mol
Then, let's divide this to the molar mass of the molecular formula.
Molar mass of M.F/Molar mass of E.F = 180/30 = 6
Therefore, let's multiply 6 to each subscript in the empirical formula to determine the actual molecular formula.
<em>Actual molecular formula = C₆H₁₂O₆</em>
Molar mass of E.F = 12 + 2(1) + 16 = 30 g/mol
Then, let's divide this to the molar mass of the molecular formula.
Molar mass of M.F/Molar mass of E.F = 180/30 = 6
Therefore, let's multiply 6 to each subscript in the empirical formula to determine the actual molecular formula.
<em>Actual molecular formula = C₆H₁₂O₆</em>