/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 How many moles of compound are i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How many moles of compound are in \(25.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) ?

Short Answer

Expert verified
0.0874 moles

Step by step solution

01

Find the Molar Mass of the Compound

To find the molar mass of \(\text{Na}_{2} \text{CO}_{3} \cdot 10 \text{H}_{2} \text{O}\), add the molar masses of all the atoms in the formula. This includes 2 sodium (Na) atoms, 1 carbon (C) atom, 3 oxygen (O) atoms for \(\text{Na}_{2} \text{CO}_{3}\), and 10 water (H\textsubscript{2}O) molecules. Use the periodic table for the atomic masses.
02

Calculate the Molar Mass

The atomic masses are: \(\text{Na}\) = 22.99 g/mol, \( \text{C} \) = 12.01 g/mol, \(\text{O}\) = 16.00 g/mol, and \(\text{H}\) = 1.01 g/mol. Calculate: \text{Na}_{2} \text{CO}_{3} = (2 x 22.99) + 12.01 + (3 x 16.00) = 105.99 g/mol Water: (10 x (2 x 1.01 + 16.00)) = 180.20 g/mol Total Molar Mass: 105.99 g/mol + 180.20 g/mol = 286.19 g/mol.
03

Convert Grams to Moles

Use the relationship: \[ \text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \]. For \(25.0 \text{ g of }\text{Na}_{2} \text{CO}_{3} \cdot 10 \text{H}_{2} \text{O}\): \[ \text{moles} = \frac{25.0 \text{ g}}{286.19 \text{ g/mol}} = 0.0874 \text{ moles} \]

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 91Ó°ÊÓ!

Key Concepts

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

molar mass
Understanding molar mass is essential in chemistry; it is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. To find the molar mass, you add up the atomic masses of all the atoms in the compound's formula. For example, let's find the molar mass of \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\) (Sodium Carbonate Decahydrate). You begin with the atomic masses: \(\text{Na}\) is 22.99 g/mol, \(\text{C}\) is 12.01 g/mol, \(\text{O}\) is 16.00 g/mol, and \(\text{H}\) is 1.01 g/mol. So, the calculation is:
\(\text{Na}_2 \text{CO}_3 : (2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99 \text{ g/mol}\).
Water: (10 \times (2 \times 1.01 + 16.00)) = 180.20 \text{ g/mol}.
Total molar mass of \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\) = 105.99 g/mol + 180.20 g/mol = 286.19 g/mol.
stoichiometry
Stoichiometry is all about the quantitative relationships in chemical reactions. It involves using ratios derived from balanced chemical equations to calculate the amounts of reactants and products. In our exercise, stoichiometry helps us understand how many moles of sodium carbonate decahydrate (\(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\)) we have when given a specific mass. The step-by-step calculation shows how stoichiometry enables the conversion of mass to moles using the molar mass as a bridge, making it a crucial tool in solving chemical equations.
moles to grams conversion
Converting between moles and grams is a fundamental skill in chemistry. In the exercise, we use the relationship: \(\text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \), to convert the given mass of \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\). For instance, for 25.0 grams of the compound, the calculation is:
\(\text{moles} = \frac{25.0 \text{ grams}}{286.19 \text{ g/mol}} = 0.0874 \text{ moles} \). This conversion is straightforward: divide the mass by the molar mass to find the number of moles. Understanding this concept is vital for accurate chemical calculations and ensures precise measurements in laboratory settings.
chemical formulas
Chemical formulas represent the types and numbers of atoms in a molecule of a compound. They provide essential information needed to compute molar masses, perform stoichiometric calculations, and understand the structure of the compound. For instance, the formula \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\) reveals that each molecule contains 2 sodium (Na) atoms, 1 carbon (C) atom, and 3 oxygen (O) atoms, plus 10 water (H\(_2\)O) molecules. Such formulas are indispensable in chemical analysis and synthesis, as they encapsulate the proportions and relationships between different atoms in a compound, helping chemists to predict reaction outcomes and to design new substances.

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

Write formulas for magnesium sulfate heptahydrate and disodium hydrogen phosphate dodecahydrate.

Sketch a heating curve for a substance \(\mathrm{X}\) whose melting point is \(40^{\circ} \mathrm{C}\) and whose boiling point is \(65^{\circ} \mathrm{C}\). (a) Describe what you will observe as a \(60 .-\mathrm{g}\) sample of \(\mathrm{X}\) is warmed from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). (b) If the heat of fusion of \(X\) is \(80 . \mathrm{J} / \mathrm{g}\), the heat of vaporization is 190 . \(\mathrm{J} / \mathrm{g}\), and if \(3.5 \mathrm{~J}\) are required to warm \(1 \mathrm{~g}\) of \(\mathrm{X}\) each degree, how much energy will be needed to accomplish the change in (a)?

A mixture of \(80.0 \mathrm{~mL}\) of hydrogen and \(60.0 \mathrm{~mL}\) of oxygen is ignited by a spark to form water. (a) Does any gas remain unreacted? Which one, \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ? (b) What volume of which gas (if any) remains unreacted? (Assume the same conditions before and after the reaction.)

How many joules of energy are required to change \(50.0 \mathrm{~g} \mathrm{Cu}\) from \(25.0^{\circ} \mathrm{C}\) to a liquid at its melting point, \(1083^{\circ} \mathrm{C}\) ? Specific heat of \(\mathrm{Cu}=0.385 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\) Heat of fusion for \(\mathrm{Cu}=134 \mathrm{~J} / \mathrm{g}\)

What are dispersion forces? Which has greater dispersion forces, \(\mathrm{C}_{3} \mathrm{H}_{8}\) or \(\mathrm{C}_{9} \mathrm{H}_{20}\) ? Justify your answer.

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.