A sample of urea contains \(1.121 \mathrm{g} \mathrm{N}, 0.161 \mathrm{g} \mathrm{H}, 0.480 \mathrm{g} \mathrm{C}\) and \(0.640 \mathrm{g}\) O. What is the empirical formula of urea?

Short Answer

Expert verified
The empirical formula of urea is \(N_2H_4CO\).

Step by step solution

01

Convert mass to moles

To find the number of moles, divide the mass of each element by its molar mass: - Nitrogen: \(1.121 \mathrm{g} \mathrm{N} \times \frac{1 \mathrm{mol} \mathrm{N}}{14.01 \mathrm{g} \mathrm{N}} = 0.080 \mathrm{mol} \mathrm{N}\) - Hydrogen: \(0.161 \mathrm{g} \mathrm{H} \times \frac{1 \mathrm{mol} \mathrm{H}}{1.008 \mathrm{g} \mathrm{H}} = 0.160 \mathrm{mol} \mathrm{H}\) - Carbon: \(0.480 \mathrm{g} \mathrm{C} \times \frac{1 \mathrm{mol} \mathrm{C}}{12.01 \mathrm{g} \mathrm{C}} = 0.040 \mathrm{mol} \mathrm{C}\) - Oxygen: \(0.640 \mathrm{g} \mathrm{O} \times \frac{1 \mathrm{mol} \mathrm{O}}{16.00 \mathrm{g} \mathrm{O}} = 0.040 \mathrm{mol} \mathrm{O}\)
02

Divide moles by smallest number of moles

The smallest number of moles among the elements is 0.040. Now divide the moles of each element by the smallest number of moles: - Nitrogen: \(\frac{0.080}{0.040} = 2\) - Hydrogen: \(\frac{0.160}{0.040} = 4\) - Carbon: \(\frac{0.040}{0.040} = 1\) - Oxygen: \(\frac{0.040}{0.040} = 1\)
03

Round ratios to nearest whole numbers

The current ratios of elements in the formula are: \(2 N : 4 H : 1 C : 1 O\) These ratios are already rounded to whole numbers.
04

Combine whole number ratios to form empirical formula

Finally, combine the whole number ratios found in step 3 to form the empirical formula: \(N_2H_4CO\) The empirical formula of urea is \(N_2H_4CO\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. Essentially, it is the calculation of reactants and products in chemical reactions. For example, when determining the empirical formula of a compound, such as urea, stoichiometry allows us to convert the mass of each element present in a sample to the number of moles. The concept empowers students to systematically approach chemical equations and balance them, relying on the law of conservation of mass which states that in a chemical reaction, matter is neither created nor destroyed.

One essential stoichiometric principle in finding an empirical formula is that the amount of each element must be converted from grams to moles since chemical reactions and formulas are based on moles rather than weight. This sets the foundation for understanding and quantifying the chemical composition of substances, as well as predicting the outcomes of chemical reactions.
Molar Mass
The molar mass of an element is defined as the mass of one mole of that element and is expressed in grams per mole (g/mol). It is a bridge between the atomic microscopic world of molecules and the macroscopic world we can measure. The molar mass tells us how many grams are in one mole of a substance and it directly corresponds to the atomic weight of an element, as seen on the periodic table.

To determine the empirical formula of a compound, you first need to calculate how many moles of each element is in your sample. This is done by dividing the mass of each element by its molar mass. For instance, carbon has a molar mass of approximately 12.01 g/mol. If you have 0.480 g of carbon, as in the urea example, you can find out the number of moles by dividing the mass (0.480 g) by its molar mass (12.01 g/mol). Understanding how to accurately calculate molar mass is crucial in stoichiometric calculations and in constructing empirical formulas.
Chemical Composition
Chemical composition refers to the identity and relative number of the elements that make up any particular compound. Every compound has a unique composition represented by a specific empirical or molecular formula. For example, determining the empirical formula of urea involves finding the simplest whole-number ratio between the atoms of the elements present.

In the urea example, the empirical formula is derived from the exact ratio of moles of each element in the sample. The chemical composition essentially determines the properties of the compound, such as reactivity, phase at room temperature, color, and biological activity. Knowledge of the chemical composition of substances is paramount in various industries, including pharmaceuticals, agriculture, and materials science. Additionally, an accurate understanding of the chemical composition is necessary for mixtures, solutions, and compounds, and is fundamental to nearly all chemical analyzes and experimental designs.
Mole to Mole Conversion
Mole to mole conversion is a key component of stoichiometry, often used to calculate the amount of reactants needed or products formed in a chemical reaction. It is based on Avogadro's number, which is approximately 6.022 x 10^23, representing the number of atoms or molecules in one mole of a substance.

In our context, after finding out the moles of each element in the compound, the next step is to divide these values by the smallest number of moles present to get a ratio. In the urea example, dividing the number of moles of nitrogen, carbon, hydrogen, and oxygen by the smallest number of moles (which was for carbon and oxygen, 0.040 moles), helps to normalize the ratio to the smallest whole numbers. Essentially, mole to mole conversions allow us to scale up from the microscopic level to macroscopic measurements, which can be practically used and measured, thus bridging the gap between the conceptual and practical aspects of chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When aluminum metal is heated with an element from Group 6A of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group 6 A element, the product is \(18.56 \%\) Al by mass. What is the formula of the compound?

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. \(\mathrm{A}\) \(752-\mathrm{g}\) sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{g}\) of pure iron by the following reaction: $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g)$$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ What masses of iron(III) oxide and aluminum must be used to produce \(15.0 \mathrm{g}\) iron? What is the maximum mass of aluminum oxide that could be produced?

Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of \(47.6 \mathrm{mg}\) cumene produces some \(\mathrm{CO}_{2}\) and \(42.8 \mathrm{mg}\) water. The molar mass of cumene is between 115 and \(125 \mathrm{g} / \mathrm{mol} .\) Determine the empirical and molecular formulas.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free