Consider sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), used in car batteries. (a) What is the molar mass of sulfuric acid? (b) What is the mass in grams of 1 mole of \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (c) What is the mass in grams of \(2.50\) moles of \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (d) What is the mass in grams of 1000 molecules of \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (Hint: Start by writing \({ }^{\prime \prime} 1000\) molecules \(\mathrm{H}_{2} \mathrm{SO}_{4}^{\prime \prime}\) and then apply conversior factors. Remember our admonition: If in doubt, convert to moles.)

Short Answer

Expert verified
The molar mass of sulfuric acid (H₂SO₄) is 98 g/mol. The mass of 1 mole of H₂SO₄ is 98 g, and the mass of 2.5 moles is 245 g. The mass of 1000 molecules of H₂SO₄ is approximately \(1.63 \times 10^{-19}\) g.

Step by step solution

01

Find the molar mass of sulfuric acid (H₂SO₄)

To find the molar mass of H₂SO₄, we need to first identify the atomic masses of hydrogen (H), sulfur (S), and oxygen (O) using the periodic table. The atomic masses are approximately: - Hydrogen (H): 1 g/mol - Sulfur (S): 32 g/mol - Oxygen (O): 16 g/mol Now we can calculate the molar mass of H₂SO₄ as follows: Molar mass of H₂SO₄ = (2 × atomic mass of H) + (1 × atomic mass of S) + (4 × atomic mass of O)
02

Calculate the molar mass of H₂SO₄

Use the atomic masses found in Step 1 to calculate the molar mass of H₂SO₄: Molar mass of H₂SO₄ = (2 × 1 g/mol) + (1 × 32 g/mol) + (4 × 16 g/mol) \(=\) 2 g/mol + 32 g/mol + 64 g/mol \(=\) 98 g/mol The molar mass of H₂SO₄ is 98 g/mol.
03

Calculate the mass of 1 mole and 2.5 moles of H₂SO₄

We can use the molar mass calculated in Step 2 to find the mass of 1 mole and 2.5 moles of H₂SO₄. a) For 1 mole of H₂SO₄: Mass = (molar mass) × (number of moles) Mass = 98 g/mol × 1 mol = 98 g b) For 2.5 moles of H₂SO₄: Mass = (molar mass) × (number of moles) Mass = 98 g/mol × 2.5 mol = 245 g
04

Calculate the mass of 1000 molecules of H₂SO₄

To calculate the mass of 1000 molecules of H₂SO₄, we first need to convert the number of molecules to moles: Number of moles = (number of molecules) / (Avogadro's number) Number of moles = 1000 molecules / \(6.022 \times 10^{23}\) molecules/mol = \(1.66 \times 10^{-21}\) mol Now we can use the molar mass to find the mass: Mass = (molar mass) × (number of moles) Mass = 98 g/mol × \(1.66 \times 10^{-21}\) mol ≈ \(1.63 \times 10^{-19}\) g So, the mass of 1000 molecules of H₂SO₄ is approximately \(1.63 \times 10^{-19}\) g.

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!

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

Butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), used as the fuel in disposable lighters, reacts with oxygen \(\left(\mathrm{O}_{2}\right)\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Suppose \(10.00 \mathrm{~g}\) butane is combined with \(10.00 \mathrm{~g} \mathrm{O}_{2}\). (a) Write a balanced chemical equation for the combustion of butane. (b) Which reactant is limiting? (c) What is the theoretical yield of each product for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) How many additional grams of the limiting reactant are required to run this reaction in a balanced fashion?

Consider the following unbalanced chemical equation: \(\mathrm{P}+\mathrm{O}_{2} \rightarrow \mathrm{P}_{2} \mathrm{O}_{5}\) (a) How many grams of phosphorus (P) are required to react completely with \(20.0 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\) (b) What is the theoretical yield in grams if you combine the amounts of reactants in part (a)?

The flavoring agent vanillin contains carbon, hydrogen, and possibly oxygen. When \(0.450 \mathrm{~g}\) of vanillin is subjected to combustion analysis, the results are \(63.08 \% \mathrm{C}\) and \(5.30 \% \mathrm{H}\). If the molar mass is approximately \(152 \mathrm{~g} / \mathrm{mol}\), what is the molecular formula of vanillin?

A \(1.000-\mathrm{g}\) sample of a liquid is subjected to combustion analysis, yielding \(92.3 \% \mathrm{C}\) and \(7.7 \% \mathrm{H}\). It may or may not also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(78 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

Nitrogen and fluorine react to form nitrogen trifluoride according to the balanced chemical equation \(\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \rightarrow 2 \mathrm{NF}_{3}(g)\) For each of the following reaction mixtures, choose the limiting reactant: (a) \(0.50 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(0.50 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (b) \(12.0 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(20.0 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (c) \(2.5 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(7.5 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (d) 100 molecules \(\mathrm{N}_{2}(g)\) and 500 molecules \(\mathrm{F}_{2}(g)\) (e) \(5.00 \mathrm{~g} \mathrm{~N}_{2}(g)\) and \(15.0 \mathrm{~g} \mathrm{~F}_{2}(g)\) (f) \(20.0 \mathrm{mg} \mathrm{N}_{2}(g)\) and \(70.0 \mathrm{mg} \mathrm{F}_{2}(g)\)

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