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What is the molecular formula of a compound with a percent composition of 49.47% C, 5.201% H, 28.84% N, and 16.47% O, and a molar mass of 194.2 g/mol?​

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Answer:

Step-by-step explanation:

The molecular formula is the true chemical formula of a covalent molecular compound, and a multiple of the empirical formula. We can use the molar mass, along with the molar empirical mass, determined from the empirical formula, to determine the molecular formula of the compound.

Using, the percentage composition of each element, we can say: In each 100g of the compound, there is 49.47g (C), 5.201g (H), 28.84g (N), and 16.47g (O).

Therefore, we can now find the number of moles of each element, in 100g of compound. number of moles (n) = mass present (m) ÷ molar mass (M). Molar mass can be found using a standard IUPAC Periodic Table.

n(C) = m/M = 49.47/12.01 = 4.1191 mol

n(H) = m/M = 5.201/1.008 = 5.1597 mol

n(N) = m/M = 28.84/14.01 = 2.0585 mol

n(O) = m/M = 16.47/16.00 = 1.0293 mol

Since the empirical formula is one in which the proportions of each element are expressed in the simplest mole ratio, therefore:

C : H : N : O = 4.1191 : 5.1597 : 2.0585 : 1.0293

Divide each ratio by 4.1191 =

1 : 1.00985 : 0.4997 : 0.249, which is approximately equal to:

1 : 1 : 0.5 : 0.25 = 2 : 2 : 1 : 0.5 = 4 : 4 : 2 : 1

Therefore, empirical formula = C₄H₄N₂O.

Using empirical formula, we can calculate molar empirical mass, which is found by summing the molar atomic weights.

molar empirical mass = 4(12.01)+4(1.008) + 2(14.01) + 16.00 = 96.092.

The molar empirical mass is double the molar mass, and therefore, molecular formula must be half the empirical formula.

Therefore, molecular formula = C₂H₂NO

User David Shuma
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