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Menthol is a flavoring agent extracted from peppermint oil. It contains C,H, and O. In one combustion analysis, 10.00mg of the substance yields 11.53mgH2O and 28.16mgCO2. What is the empirical formula of menthol?

Short Answer

Expert verified
The empirical formula of menthol is C10H20O.

Step by step solution

01

Calculate moles of CO2 produced

First, we need to find out how many moles of CO2 are produced from the combustion of 10.00 mg of menthol. The molar mass of CO2 is 44.01 g/mol. Convert the mass of CO2 to grams: 28.16mg=0.02816gCalculate moles by dividing by molar mass: moles of CO2=0.02816g44.01g/mol=6.40×104mol
02

Calculate moles of H2O produced

Now, find out how many moles of H2O are produced. The molar mass of H2O is 18.02 g/mol.Convert the mass of H2O to grams: 11.53mg=0.01153gCalculate moles by dividing by molar mass:moles of H2O=0.01153g18.02g/mol=6.40×104mol
03

Determine moles of carbon from CO2

Each mole of CO2 contains one mole of carbon. Therefore, the moles of C is equal to the moles of CO2.moles of C=6.40×104mol
04

Determine moles of hydrogen from H2O

Each mole of H2O contains two moles of hydrogen. Therefore, the moles of H is twice the moles of H2O.moles of H=2×6.40×104mol=1.28×103mol
05

Find moles of oxygen in menthol

First, determine the mass of carbon and hydrogen:- Mass of C = moles of C × molar mass of C = 6.40×104mol×12.01g/mol=7.69×103g- Mass of H = moles of H × molar mass of H = 1.28×103mol×1.008g/mol=1.29×103gMass of oxygen in menthol can be calculated by subtracting mass of C and H from total mass: mass of O=0.0100g(7.69×103g+1.29×103g)=1.02×103g
06

Calculate moles of oxygen

Convert the mass of oxygen to moles using its molar mass (16.00 g/mol):moles of O=1.02×103g16.00g/mol=6.38×105mol
07

Determine empirical formula

Now we have the moles of C, H, and O:- C: 6.40×104mol- H: 1.28×103mol- O: 6.38×105molDivide each by the smallest amount of moles (oxygen) to get the ratio:- C: 6.40×1046.38×105=10- H: 1.28×1036.38×105=20- O: 6.38×1056.38×105=1The empirical formula is C10H20O.

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Key Concepts

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

Combustion Analysis
Combustion analysis is a vital experimental procedure used to determine the composition of chemical compounds containing carbon, hydrogen, and often oxygen. The process typically involves burning a known mass of a substance and measuring the amounts of carbon dioxide (CO2) and water (H2O) produced. By capturing these products, we can deduce the content and ratio of elements such as carbon and hydrogen in the original substance. In mentoring, combustion analysis helped ascertain the quantities of both carbon and hydrogen by relating the masses of CO2 and H2O to the moles of these elements. Each molecule of CO2 represents one atom of carbon, and each H2O molecule contains two atoms of hydrogen. This straightforward relationship allows chemists to calculate the necessary ratios needed to determine the empirical formula, providing a glimpse into the chemical structure of a compound.
Molar Mass
Molar mass, also known as molecular weight, is the mass of one mole of a given substance expressed in grams per mole. This crucial concept links the macroscopic property of mass with the amount of substance measured in moles, making it fundamental in stoichiometric calculations. You can find the molar mass of any chemical compound by summing the weighted molar masses of its constituent elements.For example, in the case of menthol combustion, the molar masses of CO2 (44.01 g/mol) and H2O (18.02 g/mol) were used to convert the measured masses in milligrams to moles. This conversion was pivotal in obtaining the relative number of moles of each element involved, which was then used to determine the empirical formula. Understanding molar mass is essential in translating between the mass and amount of substance, a crucial step in accurately determining the chemical composition.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the substances involved in reactions, based on their chemical equations. It allows us to predict the amounts of reactants needed and the products formed in a chemical reaction. By using the coefficients from a balanced chemical equation, stoichiometry provides a map to navigate through the quantitative aspects of chemical transformations.When solving for the empirical formula of menthol, stoichiometry played a key role. After determining the moles of CO2 and H2O, stoichiometry was used to calculate the moles of carbon and hydrogen present in menthol. These moles were then normalized to the smallest number of moles (in this case, oxygen) to establish a simple whole-number ratio among carbon, hydrogen, and oxygen atoms, leading to the empirical formula of C10H20O.
Chemical Composition
Chemical composition refers to the identity and relative number of atoms in a substance, providing valuable information about the formula and structure of the compound. It encompasses the type and amount of each element present in a chemical species and is usually expressed as a molecular or empirical formula.The empirical formula is a fundamental concept that represents the simplest whole-number ratio of elements in a compound. In the example of menthol, its composition was deduced by calculating the relative moles of carbon, hydrogen, and oxygen derived from the product of combustion analysis. Through careful analysis and calculation, we determined that menthol’s empirical formula is C10H20O, reflecting its basic chemical makeup. Understanding chemical composition allows scientists to infer the molecular structure and potential reactivity of compounds.

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