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What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2}\), molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), molar mass \(=88 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
The molecular formulas for the given compounds are: (a) \(\mathrm{C}_{2}\mathrm{H}_{2}\mathrm{O}_{4}\) (b) \(\mathrm{C}_{4}\mathrm{H}_{8}\mathrm{O}_{2}\)

Step by step solution

01

Find the molar mass of the empirical formula

First, calculate the molar mass of the empirical formula for each compound using the atomic weights of each element. (a) For empirical formula \(\mathrm{HCO}_{2}\), the molar mass is: \[1\cdot \mathrm{m(H)} + 1\cdot \mathrm{m(C)} + 2\cdot \mathrm{m(O)} = 1.0 + 12.0 + 2\cdot 16.0 = 45.0\,\mathrm{g/mol}\] (b) For empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), the molar mass is: \[2\cdot \mathrm{m(C)} + 4\cdot \mathrm{m(H)} + 1\cdot \mathrm{m(O)} = 2 \cdot 12.0 + 4 \cdot 1.0 + 16.0 = 44\,\mathrm{g/mol}\]
02

Calculate the ratio

Find the ratio of the molar mass of the molecular formula to the molar mass of the empirical formula for each compound. This ratio is a whole number and represents the multiplication factor that you multiply the empirical formula by to reach the molecular formula. (a) For the first compound: \[\frac{\mathrm{Molar\,mass\,of\,molecular\,formula}} {\mathrm{Molar\,mass\,of\,empirical\,formula}} = \frac{90.0\mathrm{g/mol}}{45.0\mathrm{g/mol}} = 2 \] (b) For the second compound: \[\frac{\mathrm{Molar\,mass\,of\,molecular\,formula}} {\mathrm{Molar\,mass\,of\,empirical\,formula}} = \frac{88\mathrm{g/mol}}{44\mathrm{g/mol}} = 2 \]
03

Determine the molecular formula

Multiply the empirical formula by the ratio calculated in step 2 to determine the molecular formula. (a) For the first compound: \(\mathrm{HCO}_{2}\) multiplied by the ratio 2 results in the molecular formula \(\mathrm{C}_{2}\mathrm{H}_{2}\mathrm{O}_{4}\). (b) For the second compound: \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) multiplied by the ratio 2 results in the molecular formula \(\mathrm{C}_{4}\mathrm{H}_{8}\mathrm{O}_{2}\). In conclusion: (a) The molecular formula for the first compound is \(\mathrm{C}_{2}\mathrm{H}_{2}\mathrm{O}_{4}\). (b) The molecular formula for the second compound is \(\mathrm{C}_{4}\mathrm{H}_{8}\mathrm{O}_{2}\).

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Key Concepts

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

Empirical Formula
The empirical formula of a compound provides essential information about the simplest whole number ratio of atoms present. It tells you the ratio of each type of atom in a molecule, but not the actual number of atoms. For instance, the empirical formula \(\mathrm{HCO}_{2}\) suggests that hydrogen, carbon, and oxygen are present in a 1:1:2 ratio. Similarly, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) represents a ratio of 2 carbon atoms, 4 hydrogen atoms, and 1 oxygen atom.
Although an empirical formula is useful, it doesn't specify how many atoms are in a molecule, just their relative proportions. This is why determining the molecular formula, which shows the actual number of atoms, is essential for understanding a compound's real chemical structure.
The molecular formula is often a multiple of the empirical formula, depending on the actual size of the molecule concerned.
Molar Mass Calculation
Understanding how to calculate molar mass is a fundamental skill in chemistry. Molar mass is the mass of one mole of a substance and is measured in \(\mathrm{g/mol}\).
To calculate it for an empirical formula, you sum the atomic masses of all involved atoms. Using atomic weight data from the periodic table helps achieve this.
Take the empirical formula \(\mathrm{HCO}_{2}\), with atomic masses of hydrogen (\(1.0 \mathrm{g/mol}\)), carbon (\(12.0 \mathrm{g/mol}\)), and oxygen (\(16.0 \mathrm{g/mol}\)). The entire process is simple:
  • Hydrogen: \(1 \times 1.0 \mathrm{g/mol} = 1.0 \mathrm{g/mol}\)
  • Carbon: \(1 \times 12.0 \mathrm{g/mol} = 12.0 \mathrm{g/mol}\)
  • Oxygen: \(2 \times 16.0 \mathrm{g/mol} = 32.0 \mathrm{g/mol}\)
  • Total: \(1.0 + 12.0 + 32.0 = 45.0 \mathrm{g/mol}\)
This result allows you to compare empirical molar mass and given molar mass to find the actual molecular formula.
Chemical Compounds
A chemical compound is a substance made from two or more different elements chemically bonded together. Compounds have distinct properties from the elements that make them up. They form through different types of chemical bonds such as ionic or covalent bonds.

When analyzing compounds and solving problems like determining molecular formulas, understanding the structure and type of bonding can be beneficial. For instance, knowing that water (\(\mathrm{H}_2\mathrm{O}\)) has covalent bonds between hydrogen and oxygen helps in understanding its properties, such as being a polar molecule.
Compounds also come in various forms:
  • Organic Compounds: primarily composed of carbon and hydrogen, sometimes with oxygen, nitrogen, or other elements.
  • Inorganic Compounds: typically lack carbon-hydrogen bonds, and include salts like sodium chloride (\(\mathrm{NaCl}\)).
Recognizing these classifications aids in predicting the behavior and reactivity of the compounds, essential for learning chemistry.

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Most popular questions from this chapter

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

The molecular formula of allicin, the compound responsible for the characteristic smell of garlic, is \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{OS}_{2}\). (a) What is the molar mass of allicin? (b) How many moles of allicin are present in \(5.00 \mathrm{mg}\) of this substance? (c) How many molecules of allicin are in \(5.00 \mathrm{mg}\) of this substance? (d) How many \(\mathrm{S}\) atoms are present in 5.00 mg of allicin?

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} ;\) but other products containing \(\mathrm{Cl}\), such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\), are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\). (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) and \(\mathrm{HCl}\). (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily HCl: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Balance this equation. (b) Calculate the number of grams of \(\mathrm{HCl}\) that can react with \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\). (c) Calculate the number of grams of \(\mathrm{AlCl}_{3}\) and the number of grams of \(\mathrm{H}_{2} \mathrm{O}\) formed when \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Write balanced chemical equations to correspond to each of the following descriptions: (a) Solid calcium carbide, \(\mathrm{CaC}_{2}\), reacts with water to form an aqueous solution of calcium hydroxide and acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}\). (b) When solid potassium chlorate is heated, it decomposes to form solid potassium chloride and oxygen gas. (c) Solid zinc metal reacts with sulfuric acid to form hydrogen gas and an aqueous solution of zinc sulfate. (d) When liquid phosphorus trichloride is added to water, it reacts to form aqueous phosphorous acid, \(\mathrm{H}_{3} \mathrm{PO}_{3}(a q)\), and aqueous hydrochloric acid. (e) When hydrogen sulfide gas is passed over solid hot iron(III) hydroxide, the resultant reaction produces solid iron(III) sulfide and gaseous water.

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