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Calculate the molar mass of a compound if \(0.372 \mathrm{~mol}\) of it has a mass of \(152 \mathrm{~g}\).

Short Answer

Expert verified
The molar mass of the compound is approximately 408.60 g/mol.

Step by step solution

01

Understand the Relationship

The molar mass (M) of a compound represents the mass of one mole of that compound. It can be calculated using the formula: \[ M = \frac{\text{Mass of sample (in grams)}}{\text{Amount of substance (in moles)}} \]
02

Determine Known Values

From the given data, we have:- Mass of the sample: \(152 \text{ g}\)- Amount of substance: \(0.372 \text{ mol}\)
03

Apply the Formula

Substitute the known values into the formula: \[ M = \frac{152 \text{ g}}{0.372 \text{ mol}} \]
04

Calculate the Molar Mass

Perform the division:\[ M = \frac{152}{0.372} \approx 408.60 \text{ g/mol} \]
05

Verify the Units

Ensure that the resulting units are in grams per mole (g/mol), which they are, confirming the correct calculation of the molar mass.

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Key Concepts

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

Mass of Sample
The "mass of the sample" refers to the amount of matter that you're dealing with in terms of weight. When we talk about mass in chemistry, it's important to know that it's typically measured in grams. This is because grams are a part of the metric system, which is universally used in scientific work. Understanding the mass of a sample is crucial since it enables chemists to determine the amount of any given compound within a mixture. This, in turn, helps in calculating other important chemical properties, like the molar mass. Here's how you can think about it:
  • If you have a sample of a compound, and you know it weighs 152 grams, this is the sample's mass.
  • To find the molar mass of this compound, you would use this mass in your calculations.
Moreover, knowing the mass can help you ensure that you're using the correct amount of a compound for chemical reactions or experiments. In our example, the conversion from sample mass to molar mass is straightforward because we already have the mass measured in grams, making the calculation process simpler.
Amount of Substance
The "amount of substance" is a fundamental concept in chemistry, representing the number of entities, like atoms or molecules, present in a sample. This is typically measured in moles—and here's why it matters: The mole (abbreviated as mol) is a standard scientific unit for measuring large quantities of very tiny entities, such as atoms, molecules, or other specified particles. The concept is based on Avogadro's number, which is approximately 6.022 x 10²³, representing the number of atoms in 12 grams of pure carbon-12. In our exercise, the amount of substance is given as 0.372 moles. This means you have 0.372 times the number of particles contained in Avogadro's number.
  • Knowing the moles helps in comparing quantities of different substances.
  • This is especially useful when dealing with chemical reactions, where reactants and products are often measured in moles.
Understanding the amount of substance helps in determining the stoichiometry in reactions, which is crucial for balancing chemical equations and for making efficient use of chemicals in lab experiments.
Grams per Mole
The term "grams per mole" refers to molar mass, a critical concept in chemistry that provides a direct way to convert between the mass of a substance and the amount of its constituent particles. Molar mass is expressed in grams per mole (g/mol), allowing one to easily relate the mass of a chemical sample to the number of moles it contains.In the given exercise, the molar mass is calculated using the relationship:\[ M = \frac{\text{Mass of sample (in grams)}}{\text{Amount of substance (in moles)}} \]This relationship helps in finding out how much one mole of a compound weighs. For instance:
  • If a compound's sample mass is 152 grams and it contains 0.372 moles, its molar mass is found by: \[ M = \frac{152}{0.372} \approx 408.60 \text{ g/mol} \]
  • This indicates that one mole of this compound weighs approximately 408.60 grams.
With this information, chemists can accurately prepare solutions and reactants for experiments based on precise mass measurements, knowing the exact amount of substance needed for each reaction.

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Most popular questions from this chapter

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

The molar mass of caffeine is \(194.19 \mathrm{~g}\). Is the molecular formula of caffeine \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{O}\) or \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\)

Platinum forms two different compounds with chlorine. One contains 26.7 percent \(\mathrm{Cl}\) by mass, and the other contains 42.1 percent \(\mathrm{Cl}\) by mass. Determine the empirical formulas of the two compounds.

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) of mass \(13.43 \mathrm{~g}\) is completely burned in oxygen. If the total mass of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(64.84 \mathrm{~g},\) calculate the fraction of \(\mathrm{CH}_{4}\) in the mixture.

Leaded gasoline contains an additive to prevent engine "knocking." On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, "leaded gasoline"). When \(51.36 \mathrm{~g}\) of this compound is burned in an apparatus such as that shown in Figure \(3.5,55.90 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28.61 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. Determine the empirical formula of the gasoline additive. Because of its detrimental effect on the environment, the original lead additive has been replaced in recent years by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (As of \(1999,\) this compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound is burned in an apparatus like the one shown in Figure \(3.5,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of this compound?

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