Question

In a butane lighter, \(9.7 \mathrm{~g}\) of butane combine with \(34.7 \mathrm{~g}\) of oxygen to form \(29.3 \mathrm{~g}\) carbon dioxide and how many grams of water?

Short answer

15.0 grams of water are produced.

Step-by-step solution

Write the balanced chemical equation

First, determine the balanced equation for the combustion of butane (C4H10) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced equation is: 2C4H10 + 13O2 -> 8CO2 + 10H2O

Calculate moles of CO2 produced

Using the molar mass of CO2 (44.01 g/mol), calculate the moles of CO2 produced from the given mass: moles of CO2 = mass of CO2 / molar mass of CO2 = 29.3 g / 44.01 g/mol

Determine moles of H2O produced

From the balanced equation, the molar ratio of CO2 to H2O is 8:10. Thus, calculate the moles of water formed by using this ratio: moles of H2O = (moles of CO2) * (10 moles of H2O/8 moles of CO2)

Calculate mass of H2O produced

Using the molar mass of H2O (18.02 g/mol), convert the moles of H2O to grams: mass of H2O = moles of H2O * molar mass of H2O

Perform the calculations

First, calculate moles of CO2:moles of CO2 = 29.3 g / 44.01 g/mol = 0.66576 molNext, calculate moles of H2O:moles of H2O = 0.66576 mol * (10/8) = 0.83220 molFinally, calculate the mass of H2O:mass of H2O = 0.83220 mol * 18.02 g/mol = 15.0 gThis is the amount of water produced.

Key concepts

Balanced Chemical Equations

Understanding balanced chemical equations is crucial when dealing with chemical reactions. A balanced equation ensures that the number of atoms for each element is the same on both sides of the reaction, which reflects the law of conservation of mass. For instance, in the combustion of butane (C₄H₁₀), the equation must be balanced to account for all carbon (C), hydrogen (H), and oxygen (O) atoms.

To balance the equation, you would adjust the coefficients—the numbers before each compound—until there are equal numbers of each type of atom on both sides. In the example given, the balanced equation is 2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O, signifying that two molecules of butane react with thirteen molecules of oxygen to produce eight molecules of carbon dioxide and ten molecules of water. This balance is essential for calculating the quantities of reactants and products accurately in stoichiometry.

Molar Mass

The molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. The unit typically used for molar mass is grams per mole (g/mol). It tells us how much one mole of a substance weighs and is crucial for converting between the mass of a substance and the number of moles.

Every element's molar mass is found on the periodic table and is equal to its atomic weight. For compounds, the molar mass is the sum of the atomic weights of all the atoms in the molecule. For example, for carbon dioxide (CO₂), the molar mass is calculated by adding the molar mass of one carbon atom (12.01 g/mol) and two oxygen atoms (2×16.00 g/mol), which totals 44.01 g/mol. Understanding how to calculate molar mass is imperative for stoichiometry because it allows you to convert grams of a substance to moles, facilitating further stoichiometric calculations.

Moles to Grams Conversion

In stoichiometry, the conversion of moles to grams is a frequent task, and it requires the use of the molar mass as a conversion factor. To convert moles to grams, you multiply the number of moles by the molar mass of the substance.

For instance, to find out how many grams of water (H₂O) are produced in the reaction, you would use the water's molar mass (18.02 g/mol). So if you have 0.83220 moles of H₂O, the mass in grams would be calculated as follows: \(\text{mass of H}_2\text{O} = \text{moles of H}_2\text{O} \times \text{molar mass of H}_2\text{O} = 0.83220 \text{ mol} \times 18.02 \text{ g/mol}\).The result of this calculation would give you the mass of water in grams. Knowing this conversion process is key to solving stoichiometry problems, which often require converting given masses of reactants to moles and then using the balanced chemical equation to find the amounts of products formed in moles, and finally converting those moles back to grams.