Problem 29 - Introductory Chemistry Atoms First

Question

A compound is known to contain carbon and hydrogen and might also contain oxygen. A sample is burned yielding \(54.6 \% \mathrm{C}\) and \(9.16 \% \mathrm{H}\). (a) What is the empirical formula of the compound? (b) The molar mass of the compound is \(132.159 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula? (c) Write the balanced combustion reaction (reaction with \(\mathrm{O}_{2}\) ) for the compound. Answer (not worked out on purpose-you do it): (a) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}\) (c) To balance the equation, look at \(\mathrm{C}\) and \(\mathrm{H}\) first. Then balance the elemental substance \(\left(\mathrm{O}_{2}\right)\) last: We make \(\mathrm{O}_{2}\) provide \(15 \mathrm{O}\) atoms by multiplying it by \(7.5\) (that is, \(\left.\frac{15}{2}\right)\) : This is a perfectly correct balanced equation, but if you prefer the balancing coefficients to be whole mumbers, you can multiply all of them by some number that makes them whole (in this case, 2): $$ 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{3}+15 \mathrm{O}_{2} \rightarrow 12 \mathrm{CO}_{2}+12 \mathrm{H}_{2} \mathrm{O} $$

Step-by-Step Solution

Answer

(a) Empirical formula: \(C_{2}H_{4}O\) (b) Molecular formula: \(C_{6}H_{12}O_{3}\) (c) Balanced combustion reaction: \(2 C_{6}H_{12}O_{3} + 15 O_{2} \rightarrow 12 CO_{2} + 12 H_{2}O\)

Step 1: Calculate the percentage of oxygen

To find the empirical formula, first determine the percentage of oxygen present in the compound. Since there are only Carbon, Hydrogen, and Oxygen in the compound, you can simply find the percentage of oxygen by subtracting the percentage of carbon and hydrogen from 100%. Oxygen percentage = 100% - 54.6% (Carbon) - 9.16% (Hydrogen)

Step 2: Convert percentages to moles

Convert each element's percentage to moles by dividing by its atomic mass, using the values: Carbon atomic mass = 12.01 g/mol Hydrogen atomic mass = 1.01 g/mol Oxygen atomic mass = 16.00 g/mol

Step 3: Find the mole ratios

To find the mole ratios, divide the moles of each element by the smallest value among all moles. Round the resulting values to the nearest whole number.

Step 4: Write the empirical formula

Using the whole number mole ratios, write the empirical formula for the compound. #b) Molecular formula determination#

Step 5: Calculate the empirical formula mass

Find the mass of the empirical formula by adding the atomic masses of each element, multiplied by the number of atoms of each element in the empirical formula.

Step 6: Calculate n value

Divide the given molar mass by the empirical formula mass to find the value of n (n should be a whole number or closest whole number).

Step 7: Calculate the molecular formula

Multiply the empirical formula by the n value to find the molecular formula. #c) Balanced combustion reaction#

Step 8: Write the unbalanced combustion equation

Write the combustion reaction equation with the molecular formula and oxygen on the left side, and carbon dioxide and water on the right side.

Step 9: Balance the carbons and hydrogens

Balance the carbons (C) and hydrogens (H) in the equation by adjusting the coefficients of carbon dioxide (CO2) and water (H2O).

Step 10: Balance the oxygens

Balance the oxygen (O) atoms in the equation by adjusting the coefficient of oxygen (O2). If necessary, adjust other coefficients to obtain whole number coefficients.

Key Concepts

Combustion Reaction Balancing

The art of balancing a combustion reaction is crucial for understanding chemical equations and reactions. A combustion reaction typically involves a hydrocarbon (compound containing hydrogen and carbon) and oxygen, which react to form carbon dioxide and water.

In balancing these reactions, the sequence is to start with carbon and hydrogen before balancing oxygen, as oxygen is usually present in excess. It's often most convenient to leave the balancing of oxygen until the end because it can appear in multiple products and thus, its count can be manipulated more freely to achieve a balanced reaction.

For example, when balancing the combustion of a hydrocarbon, first ensure that the number of carbon atoms in the reactants is equal to the number of carbon atoms in the carbon dioxide of the products. Next, the hydrogen atoms are balanced to match the number in the water. Finally, the amount of oxygen is adjusted by tweaking the molecular oxygen (O₂) amount. If fractional coefficients appear, the entire equation can be multiplied by the smallest number that converts all coefficients to whole numbers.

Calculating Empirical Formula

The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element in the compound. It is a key step in characterizing a chemical substance.

To calculate the empirical formula, convert the percentage composition of each element to moles by dividing the percent (assuming a 100 g sample) by the atomic mass of the element. This will get the relative mol numbers for each element. Afterward, divide each of these mole values by the smallest mole number obtained. This serves to establish a common ratio. Round all the mole ratios to the nearest whole numbers to construct the empirical formula.

If you end up with a ratio that is close to a .5 value, you might have to multiply all ratios by 2 to get whole numbers. It's important that these numbers accurately reflect the simplest ratio, as they will be used to build the molecular formula next.

Molecular Formula Determination

Determining the molecular formula from the empirical formula requires the molar mass of the compound. The molecular formula is the actual number of atoms of each element in a molecule and may be a multiple of the empirical formula.

Find the mass of the empirical formula by summing the masses of all the atoms in it. Then, divide the compound's molar mass by the empirical formula mass to get a whole number, which we denote as 'n'. This number represents how many times the empirical formula must be multiplied to get the molecular formula.

For instance, if the empirical formula is CH₂O and the molar mass indicates that the molecular formula contains twice as many of each atom, then the molecular formula would be C₂H₄O₂. This process is fundamental in molecular chemistry as it unveils the exact composition of a substance.

Stoichiometry

Stoichiometry is the mathematics behind chemistry. It involves quantifying the relationships between the reactants and products in a chemical reaction based on the law of conservation of mass. The balanced chemical equation gives the stoichiometric coefficients which tell you how many moles of each reactant and product are involved.

To perform stoichiometric calculations, first, use the molar ratios from the balanced equations to convert between moles of different chemicals in the reaction. You can also calculate the mass of reactants needed or the mass of products formed. Stoichiometry enables prediction of the quantities of substances consumed and produced in a given reaction, which is critical for planning experiments and scaling reactions for industrial applications.

To ensure the most accurate stoichiometric calculations, always begin with a correctly balanced equation, verify your molar ratios, and remember to account for all possible products, especially in more complex reactions.