Question

Lactose commonly used as a binder in tablets has a molecular weight \(342 .\) What weight of \(\mathrm{CO}_{2}\) would be formed when \(1 / 12\) mole of this compound is burnt completely? \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2} \longrightarrow 12 \mathrm{CO}_{2}+11 \mathrm{H}_{2} \mathrm{O}\right)\) \((1) 12 \mathrm{~g}\) (2) \(44 \mathrm{~g}\) (3) \(4.4 \mathrm{~g}\) (4) \(440 \mathrm{~g}\)

Short answer

44 g

Step-by-step solution

Write the balanced chemical equation

Lactose (C\(_{12}\)H\(_{22}\)O\(_{11}\)) burns in the presence of oxygen (O\(_{2}\)) to produce carbon dioxide (CO\(_{2}\)) and water (H\(_{2}\)O). The balanced chemical equation is: C\(_{12}\)H\(_{22}\)O\(_{11}\) + 12 O\(_{2}\) \rightarrow 12 CO\(_{2}\) + 11 H\(_{2}\)O

Calculate moles of lactose

We have \(\frac{1}{12}\) mole of lactose. Given the molecular weight of lactose is 342 g/mol, \(\frac{1}{12}\) mole corresponds to: \(\frac{1}{12} \times 342 = 28.5 \text{ grams}\)

Determine moles of CO\(_{2}\)

From the reaction, 1 mole of lactose produces 12 moles of CO\(_{2}\). Therefore, \(\frac{1}{12}\) mole of lactose will produce: \(\frac{1}{12} \times 12 = 1 \text{ mole of CO\(_{2}\)}\)

Calculate the mass of CO\(_{2}\)

The molecular weight of CO\(_{2}\) is calculated as follows: C: 12 g/molO: 2 \times 16 = 32 g/molMolecular weight of CO\(_{2}\) = 12 + 32 = 44 g/molSince we have 1 mole of CO\(_{2}\), the mass of CO\(_{2}\) produced is: 1 mole \times 44 g/mol = 44 g

Key concepts

molecular weight

The molecular weight of a compound is the sum of the atomic weights of all the atoms in a molecule of that compound. This is also known as molecular mass or molar mass. For example, the molecular weight of lactose (C\(_{12}\)H\(_{22}\)O\(_{11}\)) is 342 g/mol. This is determined by adding the atomic masses of the constituent atoms: carbon (C, 12 atoms), hydrogen (H, 22 atoms), and oxygen (O, 11 atoms). Knowing the molecular weight is crucial, as it allows for conversions between moles and grams, which is essential in chemical calculations.

stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's a method used to relate the quantities of substances involved in a reaction. In the given problem, the stoichiometric relationship between lactose and carbon dioxide is essential. The balanced chemical equation shows that 1 mole of lactose reacts with 12 moles of oxygen to produce 12 moles of carbon dioxide and 11 moles of water. This allows us to determine how much of each product is formed based on the amount of lactose used.

chemical reactions

Chemical reactions involve the transformation of reactants into products. In this exercise, lactose burns in the presence of oxygen—a combustion reaction. The balanced equation for this reaction is: \[\text{C}_{12}\text{H}_{22}\text{O}_{11} + 12\text{O}_{2} \rightarrow 12\text{CO}_{2} + 11\text{H}_{2}\text{O}\]. This equation tells us which substances react and in what proportions. Balancing the chemical equation is critical because it ensures that the law of conservation of mass is followed—matter is neither created nor destroyed in a chemical reaction. It also provides the ratios needed for stoichiometry calculations.

mole calculations

Mole calculations are used to convert between amounts of chemical substances in grams and moles. In this problem, we need to find the mass of CO\(_{2}\) produced from the combustion of \(\frac{1}{12}\) mole of lactose. Firstly, we use the molecular weight of lactose (342 g/mol) to determine its mass for \(\frac{1}{12}\) mole, which is 28.5 grams. From the balanced chemical equation, we see that 1 mole of lactose produces 12 moles of CO\(_{2}\). Thus, \(\frac{1}{12}\) mole of lactose produces 1 mole of CO\(_{2}\). Given that the molecular weight of CO\(_{2}\) is 44 g/mol, the mass of 1 mole of CO\(_{2}\) is therefore 44 grams. These calculations show the direct relationship between moles, molecular weight, and mass.