At archery practice, Vladimir tests a bow and arrow he has recently designed. He is able to accelerate a 24.0-g arrow at an aver
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Explanation:
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In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.
(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:
And the moles of hydrogen due to the produced 1.50 grams of water:
Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:
Thus, the subscripts in the empirical formula are:
As they cannot be fractions.
(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:
Which are equal to the moles of the acid:
And the molar mass:
(3) Finally, since the molar mass of the empirical formula is:
7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol
Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:
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