Some calcium supplements contain calcium citrate, \(\mathrm{Ca}_{3}\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)_{2}\), and others contain calcium carbonate, \(\mathrm{CaCO}_{3}\). Which compound has the higher mass percent of calcium?

Short Answer

Expert verified
Calcium carbonate, CaCO3, has a higher mass percent of calcium (40.06%) compared to calcium citrate, Ca3(C6H5O7)2, which has a mass percent of 24.11%.

Step by step solution

01

Find the molar mass of calcium citrate (Ca3(C6H5O7)2)

For this, we must first find the molar mass of each element in the compound and then multiply it by the number of atoms present in the compound. Molar mass of Ca: 40.08 g/mol Molar mass of C: 12.01 g/mol Molar mass of H: 1.01 g/mol Molar mass of O: 16.00 g/mol Next, calculate the molar mass of calcium citrate: Molar mass of Ca3(C6H5O7)2 = 3(40.08) + 2(6(12.01) + 5(1.01) + 7(16.00)) = 120.24 + 2(72.06 + 5.05 + 112) = 120.24 + 2(189.11) = 120.24 + 378.22 = 498.46 g/mol
02

Find the mass percent of calcium in calcium citrate

Now that we have the molar mass of calcium citrate, we can calculate the mass percent of calcium in it. Mass percent of Ca in Ca3(C6H5O7)2 = (total mass of Ca present / molar mass of calcium citrate) * 100 = (120.24 / 498.46) * 100 = 0.2411 * 100 = 24.11 %
03

Find the molar mass of calcium carbonate (CaCO3)

We'll now follow the same procedure as in step 1 to find the molar mass of calcium carbonate. Molar mass of CaCO3 = 40.08 + 12.01 + 3(16.00) = 40.08 + 12.01 + 48.00 = 100.09 g/mol
04

Find the mass percent of calcium in calcium carbonate

With the molar mass of calcium carbonate, we can calculate the mass percent of calcium in it. Mass percent of Ca in CaCO3 = (total mass of Ca present / molar mass of calcium carbonate) * 100 = (40.08 / 100.09) * 100 = 0.4006 * 100 = 40.06 %
05

Compare the mass percent of calcium in both compounds

Finally, we compare the mass percent of calcium in both compounds: Mass percent of Ca in calcium citrate = 24.11 % Mass percent of Ca in calcium carbonate = 40.06 % Since the mass percent of calcium in calcium carbonate (40.06%) is higher than that in calcium citrate (24.11%), we can conclude that calcium carbonate contains a higher mass percent of calcium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Understanding molar mass calculation is crucial when trying to determine chemical composition and performing stoichiometric calculations. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule.

For instance, to calculate the molar mass of calcium citrate, \textbf{Ca}\(_3\)(C\(_6\)H\(_5\)O\(_7\))\(_2\), each atom's molar mass is multiplied by its occurrence in the formula: 3 atoms of calcium, 12 atoms of carbon, 10 atoms of hydrogen, and 14 atoms of oxygen. By adding these values, we get the molar mass of the compound. This step is the foundation for determining the mass percent, which represents the fraction of a particular element's mass relative to the entire compound.
Stoichiometry
Stoichiometry is the section of chemistry that involves the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of moles. To perform stoichiometric calculations, it is fundamental to understand molar ratios and the concept of the mole, which connects the microscopic world of atoms and molecules to the macroscopic world we can measure.

In our exercise, stoichiometry comes into play when calculating the mass percent of an element within a compound. The stoichiometric coefficients, which are the numbers of each element in the chemical formula, are key factors in determining the portion of each element's mass relative to the whole compound. When comparing two compounds, it's important to use stoichiometry to ensure a fair comparison regarding the amount of calcium present.
Chemical Composition
The chemical composition of a substance refers to the identity and quantity of the elements that make up that substance. Each compound has a unique chemical composition that determines its characteristics and reactions it may undergo. This composition is represented by its chemical formula, showing the types and numbers of atoms present.

Knowledge of chemical composition is vital when questioning which of two compounds contains a higher mass percent of a specific element—in this case, calcium. By determining the mass percent, we are essentially finding out how much of the compound's mass is due to calcium atoms. Since the formulas for calcium citrate and calcium carbonate contain different quantities and types of atoms, their chemical compositions directly influence the mass percent of calcium in each.

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Most popular questions from this chapter

Ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with fluorine gas \(\left(\mathrm{F}_{2}\right)\) to form carbon tetrafluoride gas and hydrogen fluoride gas. If \(2.78 \mathrm{~g}\) of ethylene reacted with an excess of fluorine, how many grams of each product could theoretically be produced?

How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would you need to get \(5.00 \times 10^{30}\) carbon atoms?

Determine the mass percent of each element in aluminum sulfate.

Butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), used as the fuel in disposable lighters, reacts with oxygen \(\left(\mathrm{O}_{2}\right)\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Suppose \(10.00 \mathrm{~g}\) butane is combined with \(10.00 \mathrm{~g} \mathrm{O}_{2}\). (a) Write a balanced chemical equation for the combustion of butane. (b) Which reactant is limiting? (c) What is the theoretical yield of each product for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) How many additional grams of the limiting reactant are required to run this reaction in a balanced fashion?

A \(1.000-\mathrm{g}\) sample of a liquid is subjected to combustion analysis, yielding \(92.3 \% \mathrm{C}\) and \(7.7 \% \mathrm{H}\). It may or may not also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(78 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

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