Write the conversion factor needed to convert from g \(\mathrm{O}_{2}\) to \(\mathrm{L} \mathrm{O}_{2}\) if the density of \(\mathrm{O}_{2}\) is 1.429 \(\mathrm{g} / \mathrm{L}\) .

Short Answer

Expert verified
The conversion factor needed to convert from g \(O_2\) to L \(O_2\) is 0.7 L/g.

Step by step solution

01

Identify the knowns

The density of \(O_2\) is given as 1.429 g/L.
02

Formulate the conversion factor

The conversion factor is simply the value that we multiply or divide by to convert one unit to another. Given the density \(density = \frac{mass}{volume}\), if we solve for volume (which in this case equals 1L), we find that the volume equals \( \frac{mass}{density}\). Thus, the conversion factor from grams to liters is \( \frac{1}{density}\).
03

Substitute the knowns

We substitute the given density of \(O_2\), 1.429 g/L, into the conversion factor, resulting in \( \frac{1}{1.429} \) L/g.
04

Evaluate

When calculated, \( \frac{1}{1.429} = 0.7\) L/g. This means that for every gram of \(O_2\), there is 0.7 liters of \(O_2\). This factor can be used to convert grams of \(O_2\) to liters of \(O_2\).

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.

Understanding Unit Conversion
Unit conversion is a fundamental aspect of chemistry and various scientific disciplines. It involves changing the representation of a quantity from one unit to another, while maintaining the same value. Take, for example, the process of converting grams of a gas to liters. This seems simple, but requires a clear understanding of the relationship between mass and volume. In chemistry, we often work with substances in different physical states, and gases, in particular, can be tricky as their volume can change with temperature and pressure.

To make these conversions, we need a conversion factor, which is a ratio that represents how a quantity in one unit can be converted to another unit. To calculate the conversion factor from grams to liters for a gas, one must use the density of the gas, which relates mass and volume. With gases, the conversion factor might also be influenced by the conditions stated in the problem, such as whether the gas is at standard temperature and pressure (STP).

In the given exercise, understanding that density provides a direct link between mass (grams) and volume (liters) is crucial. When properly applied, this conversion factor enables the transformation of mass measurements into volume measurements accurately and confidently.
Mastering Density Calculation
Density calculation is vital in chemistry for identifying substances and understanding their properties. Density is defined as mass per unit volume (\( \frac{mass}{volume} \) ) and is often expressed in grams per liter (g/L) for liquids and gases. In this exercise, the given density of oxygen gas is 1.429 g/L. This is a straightforward equation showing that for each liter of oxygen gas, the mass is 1.429 grams.

In practical terms, density tells us how concentrated a substance is. A higher density means that there is more mass packed into a given volume. By rearranging the density formula, we can solve for either mass or volume depending on the given information and what we are asked to find. Here, knowing the density allows us to find the volume of gas by dividing mass by density, which becomes the conversion factor from mass to volume.

Remember, the accuracy of such conversions depends on the precision of the density value provided and the conditions under which it was measured. In situations where temperature and pressure deviate from those under which density was determined, corrections may have to be applied.
Stoichiometry and Its Uses
Stoichiometry, in its essence, is the calculation of reactants and products in chemical reactions. It is a section of chemistry that involves using balanced equations to determine the proportions of substances involved. In problems involving gas stoichiometry, concepts like molar volume and the ideal gas law often come into play, and density can be an essential part of this as well.

In stoichiometry, unit conversion becomes pivotal. You might find yourself converting from grams to moles, moles to liters, or like in our exercise, grams to liters using the density. The stoichiometric calculations allow chemists to accurately measure out reactants for reactions and predict the amounts of products formed. These calculations are grounded in the conservation of mass principle: matter is neither created nor destroyed in a chemical reaction.

Therefore, being comfortable with concepts such as unit conversion and density calculations can greatly enhance your ability to perform stoichiometric calculations efficiently and accurately. It's like a dance of numbers and units, where each step is precisely calculated to lead to the understanding of chemical relationships and processes.

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

How many grams of Al form from 9.73 g of aluminum oxide if the yield is 91\(\% ?\) $$\mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 2 \mathrm{Al}+3 \mathrm{CO}$$

What mass of oxygen is required to burn 688 g of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) completely?

Why is a balanced chemical equation needed to solve stoichiometry problems?

Use the equation provided to answer the questions that follow. $$2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}$$ \begin{equation}\begin{array}{l}{\text { a. How many molecules of } \mathrm{H}_{2} \text { could be }} \\ {\text { made from } 27.6 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\ {\text { b. How many atoms of Na will completely }} \\ {\text { react with } 12.9 \mathrm{g} \mathrm{H}_{2} \mathrm{O} \text { ? }} \\ {\text { c. How many molecules of } \mathrm{H}_{2} \text { could form }} \\ {\text { when } 6.59 \times 10^{20} \text { atoms Na react? }}\end{array}\end{equation}

Oxygen can be prepared by heating potassium chlorate. $$2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ \begin{equation}\begin{array}{l}{\text { a. What mass of } \mathrm{O}_{2} \text { can be made from heat- }} \\ {\text { ing } 125 \text { g of } \mathrm{KClO}_{3} ?} \\ {\text { b. How many grams of } \mathrm{KClO}_{3} \text { are needed }} \\ {\text { to make } 293 \mathrm{g} \mathrm{g}_{2} \text { ? }} \\ {\text { c. How many grams of KCl could form from }} \\\ {20.8 \mathrm{g} \mathrm{KClO}_{3} ?}\end{array}\end{equation}

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