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Chapter 7: Problem 44

Glyoxyl, used in textiles; maleic acid, used to retard oxidation of fats and oils; and acontic acid, a plasticizer, all have the same empirical formula, CHO. However, the experimental molar masses are glyoxyl \(58 \mathrm{~g}\), maleic acid \(117 \mathrm{~g}\), and acontic acid \(174 \mathrm{~g}\). What are the molecular formulas of glyoxyl, maleic acid, and acontic acid?

Short Answer

Expert verified
The molecular formulas are: glyoxyl: \(C_2H_2O_2\), maleic acid: \(C_4H_4O_4\), acontic acid: \(C_6H_6O_6\).

Step by step solution

01

- Identify the empirical formula mass

First, determine the molar mass of the empirical formula CHO. The molar masses of C, H, and O are approximately 12 g/mol, 1 g/mol, and 16 g/mol respectively. Therefore, the empirical formula mass is:\[ 12 \text{ g/mol (C)} + 1 \text{ g/mol (H)} + 16 \text{ g/mol (O)} = 29 \text{ g/mol} \]
02

- Determine the molecular formula of glyoxyl

Divide the experimental molar mass of glyoxyl by the empirical formula mass to find the ratio:\[ \frac{58 \text{ g/mol}}{29 \text{ g/mol}} = 2 \]This means the molecular formula of glyoxyl is twice the empirical formula. Therefore, the molecular formula is:\[ (CHO)_2 = C_2H_2O_2 \]
03

- Determine the molecular formula of maleic acid

Similarly, divide the experimental molar mass of maleic acid by the empirical formula mass:\[ \frac{117 \text{ g/mol}}{29 \text{ g/mol}} = 4 \]The molecular formula of maleic acid is four times the empirical formula. Therefore, the molecular formula is:\[ (CHO)_4 = C_4H_4O_4 \]
04

- Determine the molecular formula of acontic acid

Divide the experimental molar mass of acontic acid by the empirical formula mass:\[ \frac{174 \text{ g/mol}}{29 \text{ g/mol}} = 6 \]The molecular formula of acontic acid is six times the empirical formula. Therefore, the molecular formula is:\[ (CHO)_6 = C_6H_6O_6 \]

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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 gives the simplest whole-number ratio of the atoms of each element in a molecule. For instance, the empirical formula CHO indicates that for every carbon atom (C), there is one hydrogen atom (H) and one oxygen atom (O). It doesn't necessarily reflect the actual numbers of atoms in the molecules, but it shows the smallest ratio between them. To find the empirical formula, you need to know the amounts of each element present in the compound and then simplify these amounts to their smallest ratio.
Molecular Mass
Molecular mass (or molecular weight) is the sum of the atomic masses of all the atoms in a molecule. Atomic masses are usually taken from the periodic table and are measured in atomic mass units (amu) or grams per mole (g/mol). For example, to find the molecular mass of the empirical formula CHO:
  • Carbon (C) has a mass of 12 g/mol
  • Hydrogen (H) has a mass of 1 g/mol
  • Oxygen (O) has a mass of 16 g/mol
Therefore, the empirical formula mass of CHO is 12 + 1 + 16 = 29 g/mol. Knowing this helps to determine the molecular formulas of compounds like glyoxyl, maleic acid, and acontic acid.
Chemistry Calculations
In chemistry, precise calculations are essential to determine various molecular properties. For example, to find the molecular formula from the empirical formula and the molar mass of a compound, you need to perform specific calculations. The formula is given by dividing the compound's molar mass by the empirical formula mass. This results in a ratio, which, when multiplied by the empirical formula, gives the molecular formula.
Let's look at glyoxyl with an empirical formula mass of 29 g/mol and a molar mass of 58 g/mol:
  • Calculate the ratio: \(\frac{58 \text{ g/mol}}{29 \text{ g/mol}} = 2\)
  • Multiply the empirical formula: \((CHO)_2 = C_2H_2O_2\)
Similar steps apply to other compounds with their respective molar masses.
Glyoxyl
Glyoxyl (also known as glyoxal) is an organic compound with the molecular formula \(C_2H_2O_2\). It has large applications in the production of textiles and as a chemical intermediate in various industries. When its molar mass, 58 g/mol, is compared to the CHO empirical formula mass of 29 g/mol, we get a ratio of 2. Thus, the molecular formula of glyoxyl is found by multiplying the empirical formula CHO by 2, resulting in \((CHO)_2\) or \(C_2H_2O_2\). Glyoxyl is a good example of how empirical formulas help in understanding molecular structures.
Maleic Acid
Maleic acid, used to retard the oxidation of fats and oils, also follows the same steps to derive its molecular formula. The molar mass of maleic acid is 117 g/mol. When divided by the empirical formula mass of 29 g/mol, it gives a ratio of 4. Hence, the molecular formula is derived as \((CHO)_4\) or \(C_4H_4O_4\). Maleic acid is often found in products that need antioxidants to extend shelf-life. These calculations help chemists develop formulas with more complex uses and properties.
Acontic Acid
Acontic acid is utilized as a plasticizer, which means it helps increase the flexibility, transparency, and durability of plastics. Its molar mass is 174 g/mol. By dividing this by the empirical formula mass of 29 g/mol, the ratio obtained is 6. This leads to the molecular formula \((CHO)_6\) or \(C_6H_6O_6\). Understanding the molecular formula is important in industries that require precise characteristics in material production. Calculations guide scientists and engineers in creating compounds with a wide array of functionalities.

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Most popular questions from this chapter

Succinic acid is \(40.7 \% \mathrm{C}, 5.12 \% \mathrm{H}\), and \(54.2 \% \mathrm{O}\). If it has an experimental molar mass of \(118 \mathrm{~g}\), what are the empirical and molecular formulas? \((7.4,7.5)\)

Iron pyrite, commonly known as "fool's gold," is \(46.5 \% \mathrm{Fe}\) and \(53.5 \% \mathrm{~S} .(7.1,7.2,7.3,7.4,7.5)\) a. If the empirical formula and the molecular formula are the same, what is the molecular formula of the compound? b. If the crystal contains \(4.85 \mathrm{~g}\) of iron, how many atoms of iron are in the crystal?

Calculate the empirical formula for each of the following: a. \(55.5 \% \mathrm{Ca}\) and \(44.5 \% \mathrm{~S}\) b. \(78.3 \%\) Ba and \(21.7 \% \mathrm{~F}\) c. \(76.0 \% \mathrm{Zn}\) and \(24.0 \% \mathrm{P}\) d. \(29.1 \% \mathrm{Na}, 40.6 \% \mathrm{~S}\), and \(30.3 \% \mathrm{O}\) e. \(19.8 \% \mathrm{C}, 2.20 \% \mathrm{H}\), and \(78.0 \% \mathrm{Cl}\)

Allyl sulfide, \(\left(\mathrm{C}_{3} \mathrm{H}_{5}\right)_{2} \mathrm{~S}\), gives garlic, onions, and leeks their characteristic odor. a. How many moles of sulfur are in \(23.2 \mathrm{~g}\) of \(\left(\mathrm{C}_{3} \mathrm{H}_{5}\right)_{2} \mathrm{~S}\) ? b. How many atoms of hydrogen are in \(0.75 \mathrm{~mol}\) of \(\left(\mathrm{C}_{3} \mathrm{H}_{5}\right)_{2} \mathrm{~S}\) ? c. How many grams of carbon are in \(4.20 \times 10^{23}\) molecules of \(\left(\mathrm{C}_{3} \mathrm{H}_{5}\right){ }_{2} \mathrm{~S}\) ? d. How many atoms of carbon are in \(15.0 \mathrm{~g}\) of \(\left(\mathrm{C}_{3} \mathrm{H}_{5}\right)_{2} \mathrm{~S}\) ?

Calculate the number of atoms of \(\mathrm{N}\) in each of the following: a. \(0.755 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) b. \(0.82 \mathrm{~g}\) of \(\mathrm{NaNO}_{3}\) c. \(40.0 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{O}\) d. \(6.24 \times 10^{23}\) molecules of \(\mathrm{NH}_{3}\) e. \(1.4 \times 10^{22}\) molecules of \(\mathrm{N}_{2} \mathrm{O}_{4}\)
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