Question

From the information below, determine the mass of substance \(C\) that will be formed if \(45.0\) grams of substance \(A\) reacts with \(23.0\) grams of substance \(B\). (Assume that the reaction between \(A\) and \(B\) goes to completion.) a. Substance \(A\) is a gray solid that consists of an alkaline earth metal and carbon ( \(37.5 \%\) by mass). It reacts with substance \(B\) to produce substances \(C\) and \(D .\) Forty million trillion formula units of \(A\) have a mass of \(4.26\) milligrams. b. \(47.9\) grams of substance \(B\) contains \(5.36\) grams of hydrogen and \(42.5\) grams of oxygen. c. When \(10.0\) grams of \(C\) is burned in excess oxygen, \(33.8\) grams of carbon dioxide and \(6.92\) grams of water are produced. \(\mathrm{A}\) mass spectrum of substance \(C\) shows a parent molecular ion with a mass-to-charge ratio of 26 . d. Substance \(D\) is the hydroxide of the metal in substance \(A\).

Short answer

The mass of substance C that will be formed is \(17.6\:g\).

Step-by-step solution

Identify elements and molar masses

: From the given information: Substance A consists of an alkaline earth metal (M) and carbon (C). Substance B consists of hydrogen (H) and oxygen (O). Substance C forms carbon dioxide (CO_2) and water (H_2O) when burned. Substance D is the hydroxide of the metal in substance A (M(OH)_2).

Find empirical formulas

: a. For substance A: The mass of 40 million trillion (40*10^12) formula units of A is 4.26 mg which is equal to 4.26*10^-3 g. The molar mass of A can be calculated as: \(Molar\:mass\:of\:A=\frac{4.26*10^-3\:g}{40*10^{12}}\) Solving, we get the molar mass of A as 106.5 g/mol. Since substance A contains 37.5% of carbon by mass, \(Mass\:of\:carbon=\frac{37.5}{100}*106.5\:g/mol\ = 39.9375\:g/mol\) The mole ratio of M to C in A is given by : \(\frac{Molar\:mass\:of\:M}{Molar\:mass\:of\:C} = \frac{106.5\:g/mole -39.9375\:g/mole}{12.01\:g/mole}\) Solving, we get the mole ratio as 5.52, which approximates to 6. Thus, the empirical formula of A is MC_6. b. For substance B: 47.9 grams of B contains 5.36 grams of H and 42.5 grams of O. Number of moles of H: \(\frac{5.36\:g}{1.008\:g/mol} = 5.32\:mol\) Number of moles of O: \(\frac{42.5\:g}{16.00\:g/mol} = 2.656\:mol\) \(Mole\:ratio\: of\: H\ to\: O\ in\: B = \frac{5.32\:moles}{2.656\:moles} = 2\) Therefore, the empirical formula of B is H_2O_2. c. For substance C: Since substance C has a parent molecular mass-to-charge ratio of 26, we know the molar mass of C is 26 g/mol. When 10 grams of C is burned, it forms 33.8 grams of CO_2 and 6.92 grams of H_2O. We can deduce the empirical formula of C by calculating the moles of carbon and hydrogen: Moles of carbon in CO_2: \(\frac{33.8\:g}{44.01\:g/mol} = 0.768\:mol\) Moles of hydrogen in H_2O: \(\frac{6.92\:g}{18.015\:g/mol} = 0.384\:mol\) From these values, we see that the mole ratio of carbon to hydrogen in substance C is 2:1. Hence, the empirical formula of C is CH_2.

Determine balanced chemical equation

: Now we have the following equation: \(MC_6 + H_2O_2 \rightarrow CH_2 + M(OH)_2\) To balance the equation: \(3MC_6 + 6H_2O_2 \rightarrow 6CH_2 + 3M(OH)_2\)

Calculate limiting reactant

: Given, 45.0 grams of A and 23.0 grams of B. Calculate the moles of A and B: Moles of A: \(\frac{45.0\:g}{106.5\:g/mol} = 0.422\:mol\) Moles of B: \(\frac{23.0\:g}{34.014\:g/mol} = 0.676\:mol\) The stoichiometric ratio for A to B from the balanced equation: \(\frac{Moles\:of\:A}{Moles\:of\:B} = \frac{0.422\:mol}{0.676\:mol} = 0.625\) Comparing the stoichiometric ratio to the actual mole ratio from the balanced equation: \(\frac{Moles\:of\:A}{Moles\:of\:B}=\frac{3\:mol}{6\:mol} = 0.5\) Since 0.625 > 0.5, A is in excess, and B is the limiting reactant.

Compute mass of substance C

: Since B is the limiting reactant, we can now calculate the mass of substance C produced by the reaction. Using the balanced equation: 6 mol of C produced from 6 mol of B, so moles of C formed = 0.676 mol. Mass of C: \(0.676\:mol * 26\:g/mol = 17.576\:g\) Therefore, the mass of substance C that will be formed is \(17.6\:g\).

Key concepts

Chemical Reactions

When we talk about chemical reactions, we're referring to a process in which substances, known as reactants, are transformed into different substances, referred to as products. This transformation involves breaking chemical bonds in reactants and forming new bonds in products. A simple way to represent this is with a chemical equation, which shows the reactants on the left side and the products on the right side. This process is critically governed by the law of conservation of mass, stating that mass cannot be created or destroyed in a chemical reaction. Therefore, the number of atoms for each element should be balanced on both sides of the equation.

In the exercise given, substance A reacts with substance B to form substances C and D. Once we identify the chemical composition of these substances, the next step involves writing a balanced chemical equation. Balancing the equation ensures that all atoms are accounted for in the reaction, maintaining the balance dictated by the conservation law. This balance helps predict the quantities of products formed from a certain amount of reactants.

Limiting Reactants

In any chemical reaction, the reactant that gets completely consumed first and thus limits the extent of the reaction is known as the limiting reactant. Determining the limiting reactant is crucial as it determines the maximum amount of products that can be formed. We find the limiting reactant by comparing the ratio of moles of each reactant used to the ratio required by the balanced chemical equation.

Once you have a balanced equation, the stoichiometric coefficients tell you how the quantities of the reactants relate to each other. In the example given, substance A and substance B react to form products. By calculating the actual moles of each reactant and comparing them to the balanced equation's stoichiometric ratio, we can figure out which reactant is in excess and which one is the limiting reactant.

In this case, the calculations reveal that substance B is the limiting reactant. This means that the amount of product formed, substance C, is directly dependent on the availability of substance B, since all of it will be used before substance A is fully consumed.

Empirical Formulas

Empirical formulas provide the simplest whole-number ratio of atoms of each element in a compound. Unlike molecular formulas, which show the exact number of atoms of each element, empirical formulas give the simplest possible ratio. To derive an empirical formula, we start by converting the masses of each element to moles and then determine the simplest whole-number ratio.

In the exercise at hand, we got the empirical formulas of substances based on their elemental compositions. For instance, substance A's empirical formula is derived from its composition of an alkaline earth metal and carbon. Similarly, substance B is composed of hydrogen and oxygen, leading to its empirical formula. Calculating and using empirical formulas are essential steps in stoichiometry, providing the groundwork for understanding the possible chemical reactions and calculating the masses of products and reactants involved.

Empirical formulas are particularly important when determining chemical equations' stoichiometry, as they help chemists predict reactant consumption and product formation in reactions.