Question

Consider the following balanced chemical equation: $$ A+5 B \longrightarrow 3 C+4 D $$ a. Equal masses of \(\mathrm{A}\) and \(\mathrm{B}\) are reacted. Complete each of the following with either "A is the limiting reactant because \(" ;{ }^{\prime *} \mathrm{~B}\) is the limiting reactant because \("\) or "we cannot determine the limiting reactant because i. If the molar mass of \(\mathrm{A}\) is greater than the molar mass of B, then ii. If the molar mass of \(\mathrm{B}\) is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound \(\mathrm{A}\) has the same molar mass as carbon dioxide. Compound \(\mathrm{B}\) is a diatomic molecule. Identify \(\mathrm{com}-\) pound \(\mathrm{B}\) and support your answer. c. Compound \(\mathrm{A}\) is a hydrocarbon that is \(81.71 \%\) carbon by mass. Detemine its empirical and molecular formulas.

Short answer

a. i. A is the limiting reactant because we cannot form more products when A is completely consumed. ii. B is the limiting reactant because we cannot form more products when B is completely consumed. b. Compound B is diatomic oxygen (O₂). c. The empirical formula of compound A is CH₃, and the molecular formula is C₃H₉.

Step-by-step solution

(Question a, part i) Determine the limiting reactant when the molar mass of A is greater than B.

Since we are given equal masses of A and B, we need to consider their molar ratios to determine the limiting reactant. First, we'll compare the molar mass. Given that the molar mass of A is greater than the molar mass of B, then the number of moles of A in equal masses will be less than the moles of B. Therefore, A is the limiting reactant because we cannot form more products when A is completely consumed.

(Question a, part ii) Determine the limiting reactant when the molar mass of B is greater than A.

In the case where the molar mass of B is greater than the molar mass of A, the number of moles of B in equal masses will be less than the moles of A. Therefore, B is the limiting reactant because we cannot form more products when B is completely consumed.

(Question b) Identify compound B

In the chemical equation, compound C is carbon dioxide (CO₂), and compound D is water (H₂O). Compound A has the same molar mass as CO₂, which means that the molar mass of A is \(44.01 g/mol\). Compound B is a diatomic molecule. To identify compound B, we'll look at the balanced chemical equation: \(A + 5B \longrightarrow 3C + 4D\). Since A is a hydrocarbon, it will only contain carbon (C) and hydrogen (H). The balanced chemical equation suggests that compound B must be reactant which could provide ample oxygen (O) to form CO₂ and H₂O. So B is diatomic oxygen, O₂.

(Question c, part i) Determine the empirical formula of compound A

We are given that compound A is a hydrocarbon and is \(81.71\%\) carbon by mass. Therefore, the remaining mass percentage in compound A is hydrogen, which is \(100\%-81.71\%=18.29\%\). To determine the empirical formula, we'll assume we have 100 grams of the compound, so: - Carbon: \(81.71 g\) (mass) × \(\frac{1 mol}{12.01 g}\) (molar mass) = \(6.808 mol\) C - Hydrogen: \(18.29 g\) (mass) × \(\frac{1 mol}{1.008 g}\) (molar mass) = \(18.148 mol\) H Divide the moles of each element by the smaller of the two mole values: - Carbon: \(\frac{6.808}{6.808}\) = 1 - Hydrogen: \(\frac{18.148}{6.808}\) ≈ 2.67 Since it is close to 3, we can approximate it to get the empirical formula: \(\mathrm{CH_3}\).

(Question c, part ii) Determine the molecular formula of compound A

Since compound A has the same molar mass as CO₂, \(44.01 g/mol\), we need to determine the multiple of the empirical formula mass which will match this molar mass. First, let's calculate the empirical formula mass: Empirical formula mass = \(12.01 (C) + 3(1.008 (H)) = 12.01 + 3.024 = 15.034 g/mol\) Now, we'll determine the multiple of the empirical formula mass that matches the molar mass: Molar mass multiple = \(\frac{44.01}{15.034}\) ≈ 2.92, which we can approximate to 3. Therefore, the molecular formula of compound A is 3 times the empirical formula: Molecular formula = \(\mathrm{C_3H_{9}}\)

Key concepts

Molar Mass

Molar mass is an essential concept in chemistry, helping us compare different substances by their mass. It is defined as the mass of one mole of a given substance and is typically expressed in grams per mole (g/mol). Additionally, the molar mass of an element can be found by looking at the periodic table, where it corresponds to the atomic weight for single elements. For compounds, molar mass is the sum of the atomic masses of all constituent atoms in the chemical formula.

Understanding molar mass allows us to convert between mass and moles, which helps in determining the limiting reactant. In the exercise, we were tasked with deducing the limiting reactant based on the molar mass of different substances. If the molar mass of substance A is greater than that of substance B, fewer moles of A will be present compared to B, assuming equal masses. Thus, identifying which substance is available in lesser moles becomes crucial to finding which one will limit the reaction's progression.

Balanced Chemical Equation

A balanced chemical equation is crucial for understanding chemical reactions. It ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the conservation of mass principle.

In the exercise, the reaction is represented by a balanced chemical equation: \( A + 5B \longrightarrow 3C + 4D \). Here, it specifies that one mole of A reacts with five moles of B to produce three moles of C and four moles of D. This equation is key for determining the ratios of reactants and products. It implies that even if equal masses are taken, the actual required amounts depend on these ratios. When developing any stoichiometry calculations, having a balanced equation allows chemists to accurately predict how much of each substance is needed or produced, and to identify the limiting reactant, thereby efficiently using resources in a laboratory setting.

Empirical Formula

The empirical formula of a compound provides the simplest whole-number ratio of the elements present in that compound. It gives us an idea about the relative quantities of each type of atom in the compound, but not the actual number of atoms.

In the exercise, we found that compound A is a hydrocarbon with 81.71% carbon by mass. To determine the empirical formula, we proceed by calculating the moles of each element based on the percentage composition. For carbon, this results in roughly the equivalent moles as we found for hydrogen. However, because hydrogen was calculated to have a higher relative amount, leading us to derive an approximate ratio that gives an empirical formula of \( \text{CH}_3 \). This provides a basic representation of the compound's composition, valuable for identifying unknown substances and in chemical synthesis.

Molecular Formula

The molecular formula of a compound reveals the actual number of atoms of each element present in a molecule of that compound. It is distinct from the empirical formula, which shows only the simplest ratio.

In our exercise, to ascertain the molecular formula of compound A, we first understood the empirical formula \( \text{CH}_3 \) and the compound's molar mass of \( 44.01 \text{ g/mol} \). By comparing this to the empirical formula's mass \( 15.034 \text{ g/mol} \), we found a multiplication factor of approximately 3. Multiplying the empirical formula by this factor provided the molecular formula \( \text{C}_3\text{H}_9 \). Knowing both the empirical and molecular formulas is important as the molecular formula offers details about the actual construct of the molecule, which is crucial in fields like organic chemistry, where the structure deeply influences properties and reactivity.