Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Tooth enamel is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH}) .\) Calculate the percent composition of the elements present.

Short Answer

Expert verified
Calcium: 41.21%, Phosphorus: 19.10%, Oxygen: 39.49%, Hydrogen: 0.21%.

Step by step solution

01

Identify the Molar Mass

Identify the molar mass of each element in the chemical formula \( \mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH}) \). The atomic masses are approximately: \( \mathrm{Ca}: 40.08 \, \mathrm{g/mol} \), \( \mathrm{P}: 30.97 \, \mathrm{g/mol} \), \( \mathrm{O}: 16.00 \, \mathrm{g/mol} \), and \( \mathrm{H}: 1.01 \, \mathrm{g/mol} \).
02

Calculate the Molar Mass of Each Element in the Compound

Calculate the total molar mass of each element in the compound:- \(5 \times 40.08 = 200.40 \, \mathrm{g/mol} \) for calcium- \(3 \times 30.97 = 92.91 \, \mathrm{g/mol} \) for phosphorus- \( (3 \times 4 + 1) \times 16.00 = 192.00 \, \mathrm{g/mol} \) for oxygen- \(1.01 \, \mathrm{g/mol} \) for hydrogen.
03

Calculate the Total Molar Mass of the Compound

Sum the molar masses of all elements to obtain the total molar mass:\[200.40 + 92.91 + 192.00 + 1.01 = 486.32 \, \mathrm{g/mol}.\]
04

Determine the Percent Composition of Each Element

Calculate the percent composition of each element:- Calcium: \((200.40 / 486.32) \times 100 \approx 41.21\%\)- Phosphorus: \((92.91 / 486.32) \times 100 \approx 19.10\%\)- Oxygen: \((192.00 / 486.32) \times 100 \approx 39.49\%\)- Hydrogen: \((1.01 / 486.32) \times 100 \approx 0.21\%\).

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.

Molar Mass Calculation
When working with chemical formulas, understanding molar mass is crucial. Molar mass is essentially the mass of one mole of a substance and is expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you need to know the atomic masses of the individual elements present.
  • The atomic mass for calcium (\( \mathrm{Ca} \)) is 40.08 g/mol.
  • For phosphorus (\( \mathrm{P} \)), it's 30.97 g/mol.
  • Oxygen (\( \mathrm{O} \)) has an atomic mass of 16.00 g/mol.
  • While hydrogen (\( \mathrm{H} \)) is 1.01 g/mol.
To find the compound's molar mass, multiply the atomic masses by their respective counts in the molecule, and sum them up. For example, in tooth enamel, we're dealing with the formula \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})\.\)
Here, you multiply the molar mass of each element by its corresponding number of atoms, and add these to get the total molar mass of 486.32 g/mol.
Tooth Enamel Chemistry
The chemistry of tooth enamel is fascinating and integral to understanding its composition. Tooth enamel is primarily made of a compound known as hydroxyapatite, represented by the chemical formula \( \mathrm{Ca}_{5}(\mathrm{PO}_{4})_{3}(\mathrm{OH})\)\. This compound is a form of calcium phosphate and is responsible for the hardness of tooth enamel.
Enamel mainly consists of calcium, phosphorous, oxygen, and hydrogen. These elements come together to form a strong and resilient outer layer for our teeth.
  • Calcium contributes to the structural integrity of enamel.
  • Phosphorus helps in building strength and resistance against decay.
  • Oxygen and hydrogen bond within the hydroxide ions (\( \mathrm{OH}^{-} \)) in the compound, playing significant roles in the stability of the compound network.
Recognizing these elements and their contributions offers insights into oral health and dental care.
Elemental Composition Calculation
Elemental composition calculation tells us the proportion of each element in a compound. This is done by determining the percentage of each element with respect to the total molar mass of the compound. The step-by-step approach involves calculating the mass of individual elements in the compound and dividing it by the total molar mass.
For instance:
  • The percent composition for calcium is found by taking its total mass in the compound (\(200.40\ g/mol\)) and dividing it by the total molar mass (\(486.32\ g/mol\)), then multiplying by 100. This gives roughly 41.21\%.
  • Similarly, for phosphorus, oxygen, and hydrogen, the calculations yield 19.10\%, 39.49\%, and 0.21\% respectively.
Calculating the percent composition is a valuable tool in chemistry as it helps us understand the makeup of compounds and points towards their potential characteristics and functionality. It's especially useful in fields like material science and biochemistry where the elemental makeup is directly related to the substance's properties and applications.

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

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2} $$ Suppose that \(1.64 \times 10^{3} \mathrm{~kg}\) of Fe is obtained from a \(2.62 \times 10^{3}-\mathrm{kg}\) sample of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Assuming that the reaction goes to completion, what is the percent purity of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the original sample?

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a component of gasoline. Complete combustion of octane yields \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\). Incomplete combustion produces \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO},\) which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gallon (gal) of octane is burned in an engine. The total mass of \(\mathrm{CO}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(11.53 \mathrm{~kg} .\) Calculate the efficiency of the process; that is, calculate the fraction of octane converted to \(\mathrm{CO}_{2}\). The density of octane is \(2.650 \mathrm{~kg} / \mathrm{gal}\).

An impure sample of zinc \((\mathrm{Zn})\) is treated with an excess of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) to form zinc sulfate \(\left(\mathrm{ZnSO}_{4}\right)\) and molecular hydrogen \(\left(\mathrm{H}_{2}\right) .\) (a) Write a balanced equation for the reaction. (b) If \(0.0764 \mathrm{~g}\) of \(\mathrm{H}_{2}\) is obtained from \(3.86 \mathrm{~g}\) of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in part ( \(b\) )?

Consider the combustion of butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l) $$ In a particular reaction, \(5.0 \mathrm{~mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10}\) react with an excess of \(\mathrm{O}_{2}\). Calculate the number of moles of \(\mathrm{CO}_{2}\) formed.

Calculate the mass in grams of iodine \(\left(\mathrm{I}_{2}\right)\) that will react completely with \(20.4 \mathrm{~g}\) of aluminum (Al) to form aluminum iodide \(\left(\mathrm{AlI}_{3}\right)\).

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