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Chapter 7: Problem 9

How is molar mass of an element used to convert from number of moles to mass in grams?

Short Answer

Expert verified
Molar mass, which is equivalent to the atomic weight in grams, is used to convert from number of moles to mass in grams by multiplication. The formula used is: Mass in grams = Number of moles * Molar mass.

Step by step solution

01

Understand the Concept of a Mole

Firstly it is important to know that a mole is a unit of measurement used in chemistry to express the amount of a chemical substance. It helps to count entities at a microscopic level.
02

Learn About Molar Mass

Next step is to understand Molar mass. It is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. The molar mass of an element is numerically equal to the element's atomic weight in g/mol. For example, the atomic weight of oxygen is approximately \(16 u\), therefore, its molar mass is \(16 g/mol\).
03

Conversion from Moles to Grams

For converting an amount from moles to grams, we use the formula: Mass in grams = number of moles * molar mass. This formula demonstrates that the number of moles is directly proportionate to the mass in grams.

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Most popular questions from this chapter

Oleic acid has the empirical formula \(\mathrm{C}_{9} \mathrm{H}_{17} \mathrm{O} .\) If the experimental molar mass is \(282 \mathrm{g} / \mathrm{mol},\) what is the molecular formula of oleic acid?

Why is the ratio between the empirical formula and the molecular formula a whole number?

How many atoms of Fe are in the formula \(\mathrm{Fe}_{3} \mathrm{C} ?\) How many moles of Fe are in one mole of \(\mathrm{Fe}_{3} \mathrm{C}\) ?

What particular isotope is the basis for defining the atomic mass unit and the mole?

How many moles of compound are in each of the following samples: $$ \begin{array}{l}{\text { a. } 6.60 \mathrm{g}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \text { ( molar mass of }} \\\ {\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}=132.17 \mathrm{g} / \mathrm{mol} )} \\ {\text { b. } 4.5 \mathrm{kg} \text { of } \mathrm{Ca}(\mathrm{OH})_{2} \text { (molar mass of }} \\\ {\mathrm{Ca}(\mathrm{OH})_{2}=74.10 \mathrm{g} / \mathrm{mol} )}\end{array} $$
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