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Chapter 9: Problem 96

Determine the empirical formula of the compound with the following mass percents of the elements present: \(58.5 \% \mathrm{C} ; 4.91 \% \mathrm{H} ; 19.5 \% \mathrm{O} ; 17.1 \% \mathrm{~N}\).

Short Answer

Expert verified
The empirical formula of the compound with the given mass percentages is C₄H₄NO.

Step by step solution

01

Assume a 100g sample

Assuming a 100g sample allows us to directly use the given percentages as masses for each element, as the percentages are intended to represent the mass of each element in the compound.
02

Convert mass to moles for each element

Using the atomic masses from the periodic table, we will convert the mass of each element to moles: Carbon: \(58.5g C \cdot \frac{1\, mol\, C}{12.01\, g\, C} = 4.87\, mol\, C\) Hydrogen: \(4.91g H \cdot \frac{1\, mol\, H}{1.01\, g\, H} = 4.86\, mol\, H\) Oxygen: \(19.5g O \cdot \frac{1\, mol\, O}{16.00\, g\, O} = 1.22\, mol\, O\) Nitrogen: \(17.1g N \cdot \frac{1\, mol\, N}{14.01\, g\, N} = 1.22\, mol\,N\)
03

Determine the smallest number of moles

Identify the smallest number of moles among all the elements calculated in the previous step: Smallest value = 1.22 mol
04

Calculate the mole ratios

Divide the moles of each element by the smallest value to find the mole ratios: Carbon: \(\frac{4.87\, mol\, C}{1.22\, mol} = 3.99 \approx 4\) Hydrogen: \(\frac{4.86\, mol\, H}{1.22\, mol} = 3.98 \approx 4\) Oxygen: \(\frac{1.22\, mol\, O}{1.22\, mol} = 1\) Nitrogen: \(\frac{1.22\, mol\, N}{1.22\, mol} = 1\)
05

Write the empirical formula

Combine the mole ratios calculated in the previous step as subscripts for each element to write the empirical formula: Empirical formula: C₄H₄NO

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Most popular questions from this chapter

A \(1.000-g\) sample of a liquid that contains only carbon and hydrogen burns in oxygen to produce \(1.284 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What are the mass percents of the elements present in this sample? (b) What is the empirical formula for this compound? (c) The molar mass of this compound is determined to be about \(71 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound? (Hint: When attempting this problem, understand that all of the carbon in the compound burned ends up as \(\mathrm{CO}_{2}\), and all of the hydrogen in the compound burned ends up as \(\mathrm{H}_{2} \mathrm{O}\). Also, there is only 1 mole of \(C\) per mole of \(\mathrm{CO}_{2}\), but there are 2 moles of \(\mathrm{H}\) per mole of \(\mathrm{H}_{2} \mathrm{O} .\) )

Sucrose has the molecular formula \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} .\) If you were to completely burn \(2.00 \mathrm{~g}\) of sucrose in a stream of oxygen, how many grams of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) would be produced?

Consider the unbalanced chemical equation \(\mathrm{Cl}_{2} \mathrm{O}_{7}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HClO}_{4}\) The reaction is carried out at \(82.0 \%\) yield and gives \(52.8 \mathrm{~g}\) of \(\mathrm{HClO}_{4}\) (a) What is the theoretical yield of \(\mathrm{HClO}_{4}\) ? (b) How many grams of \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) and \(\mathrm{H}_{2} \mathrm{O}\) were consumed in the reaction?

A compound used as an insecticide that contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{Cl}\) is subjected to combustion analysis, yielding \(55.55 \% \mathrm{C}\) and \(3.15 \% \mathrm{H}\). (a) What is the empirical formula of this compound? (b) What is the molecular formula if its empirical formula is \(1 / 3\) the mass of its actual formula?

If you have \(200.0 \mathrm{~g}\) of dinitrogen pentoxide, how many atoms of nitrogen do you have? How many atoms of oxygen do you have?
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