(i) Butyric acid contains only \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\). A \(4.24 \mathrm{mg}\) sample of butyric acid is completely burned. It gives \(8.45 \mathrm{mg}\) of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and \(3: 46 \mathrm{mg}\) of water. What is the mass percentage of each element in butyric acid? (ii) If the elemental composition of butyric acid is found to be \(54.2 \%\) C, \(9.2 \% \mathrm{H}\) and \(36.6 \%\) O, determine the empirical formula. (iii) The molecular mass of butyric acid was determined by experiment to be \(88 .\) What is the molecular formula?

Short Answer

Expert verified
Butyric acid consists of 53.77% Carbon, 10.61% Hydrogen, and 35.62% Oxygen. The empirical formula is C4H8O2, and the molecular formula is C4H8O2 since the empirical formula mass matches the molecular mass of 88.

Step by step solution

01

Determine masses of elements from combustion products

Calculate the mass of carbon from the mass of CO2 produced, using the ratio of the molar mass of C to the molar mass of CO2. Similarly, calculate the mass of hydrogen from the mass of H2O produced, using the ratio of the mass of H to the molar mass of H2O.
02

Calculate mass percentage of each element in butyric acid

Divide the masses of carbon and hydrogen by the original mass of butyric acid and multiply by 100 to find the mass percentage of C and H. The mass percentage of O is found by subtracting the percentage of C and H from 100.
03

Derive empirical formula from mass percentages

Determine moles of each element using their mass percentages and atomic masses. Divide the mole values by the smallest number of moles to find the mole ratio of C, H, and O. This ratio provides the subscripts for the empirical formula.
04

Determine molecular formula from empirical formula and molecular mass

Calculate the empirical formula mass. Then, determine the number of empirical units in the molecular formula by dividing the molecular mass by the empirical formula mass. Multiply the subscripts in the empirical formula by this number to find the molecular formula.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Analysis
Combustion analysis is a chemical process used to determine the elemental composition of a compound by fully burning the sample in excess oxygen and analyzing the products. In the case of butyric acid, when it burns, it produces carbon dioxide (CO2) and water (H2O). From the respective masses of these products, we can calculate the amounts of carbon and hydrogen in the original compound. For example, each molecule of CO2 contains one carbon atom, so by finding out how many moles of CO2 are produced, we can determine the moles of carbon in the butyric acid. The same principle applies to the calculation of hydrogen using the produced water, since each molecule of H2O contains two hydrogen atoms.

This method relies on the principle of mass conservation: the total mass of the reactants equals the total mass of the products. In a well-planned combustion analysis, any nitrogen present would form nitrogen gas (N2), and sulfur would form sulfur dioxide (SO2), which can also be measured if those elements were present in the original sample.
Mass Percentage Calculation
The mass percentage of each element in a compound represents the fraction of the total mass that is attributed to that specific element. It is calculated by dividing the mass of the element in the compound by the total mass of the compound, then multiplying by 100 to convert the fraction into a percentage. For our analysis, once we know the mass of carbon from the CO2 and the mass of hydrogen from the H2O, we simply divide these by the original sample mass of butyric acid and multiply by 100.

To find the mass percentage of oxygen, it isn't necessary to measure it directly; we can subtract the sum of the mass percentages of carbon and hydrogen from 100%. This approach takes advantage of the fact that the compound only contains carbon, hydrogen, and oxygen, ensuring no mass is from any other element.
Empirical Formula Determination
The empirical formula is a simple expression that shows the smallest whole number ratio of the elements in a compound. To find it, we start with the mass percentages of elements and convert those to moles, using their atomic masses. This step involves dividing the mass percentage by the atomic mass for each element, which gives us the respective mole ratios.

Once we have the mole ratios, we divide each by the smallest of the mole ratios to normalize them to the smallest possible whole numbers. These normalized ratios can sometimes be fractional, and in such cases, we multiply all ratios by the smallest number that converts all of them into whole numbers. These whole numbers are the subscripts used in the empirical formula of the compound, representing the relative numbers of atoms of each element present.
Molecular Formula Calculation
The molecular formula of a compound is a multiple of its empirical formula and shows the actual number of atoms of each element in a molecule of the compound. To determine the molecular formula, we need the empirical formula and the molecular mass of the compound, which is experimentally determined.

First, we calculate the mass of the empirical formula (empirical formula mass) by summing up the atomic masses of all atoms in the empirical formula. Then, we find the ratio of the molecular mass to the empirical formula mass. This ratio, which is usually a whole number, tells us how many empirical formula units are in the actual molecule. We multiply the subscripts in the empirical formula by this number to get the molecular formula of the compound. For butyric acid, if the empirical formula unit had a certain number of carbon, hydrogen, and oxygen atoms, and it was determined that the molecular mass was 88, then we'd use this information to deduce the actual number of each type of atom in a single molecule of butyric acid.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free