How many \(g\) atoms of \(S\) are present in \(4.9 \mathrm{~g} \mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short Answer

Expert verified
There are approximately 1.60 g of S atoms in 4.9 g of \(H_2SO_4\).

Step by step solution

01

Calculate the molar mass of \(H_2SO_4\)

To calculate the molar mass of \(H_2SO_4\), sum the atomic masses of all the atoms in one molecule of sulfuric acid. The atomic masses (in grams per mole) are approximately 1.01 for hydrogen (H), 32.07 for sulfur (S), and 16.00 for oxygen (O). There are two hydrogens, one sulfur, and four oxygens in \(H_2SO_4\), so the molar mass is \(2(1.01) + 32.07 + 4(16.00) = 2.02 + 32.07 + 64.00 = 98.09 \text{g/mol}\).
02

Determine the moles of \(H_2SO_4\) in 4.9 g

Use the molar mass of \(H_2SO_4\) to convert the mass in grams to moles. The number of moles is equal to the mass of the substance (in grams) divided by its molar mass (in grams per mole): \[ \text{moles of} \(H_2SO_4\) = \frac{4.9 \text{g}}{98.09 \text{g/mol}} \approx 0.04997 \text{mol}. \]
03

Calculate the grams of Sulfur (S) in \(H_2SO_4\)

Since there is one S atom in each molecule of \(H_2SO_4\), the number of moles of S is equal to the number of moles of \(H_2SO_4\). To find the mass in grams of S, multiply the moles of S by the atomic mass of S (32.07 g/mol): \[ \text{grams of S} = 0.04997 \text{mol} \times 32.07 \text{g/mol} = 1.60 \text{g}. \]

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Key Concepts

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

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. Think of it as a way of balancing a chemical budget, ensuring that the amount of substances going in matches what comes out.

When using stoichiometry in calculations, like the exercise we're examining, it's crucial to start with a balanced chemical equation. It gives you the ratios of how much of one substance reacts with another or is produced. This exercise didn't require balancing equations since we're only concerned with the composition of a single substance. However, the underlying principle of stoichiometry—that matter is conserved in a reaction and everything must be accounted for—still applies.
Mole Concept
The mole concept is an essential part of chemistry that helps us count small particles like atoms, ions, and molecules using Avogadro's number \(6.022 \times 10^{23}\). A mole represents a standard quantity—the same number of units as there are atoms in exactly 12 grams of pure carbon-12.

When we say a substance has a molar mass of 98.09 g/mol, like \(H_2SO_4\) in our exercise, we're saying that one mole of \(H_2SO_4\) weighs 98.09 grams. By converting mass to moles or vice versa, we can relate our macroscopic experiments to the actions of atoms and molecules at the microscopic level.
Atomic Mass
Atomic mass is the mass of a single atom, usually expressed in atomic mass units (amu), where carbon-12 is defined as having an exact mass of 12 amu. This concept extends to 'grams per mole' when used in bulk calculations.

Each atom on the periodic table has a unique atomic mass, reflecting the combined number of protons and neutrons in its nucleus. For example, sulfur (S) has an atomic mass of roughly 32.07 g/mol. In the given exercise, knowing the atomic mass allows us to find the mass of sulfur in a compound by multiplying it with the number of moles of that element.
Chemical Formula
A chemical formula, such as \(H_2SO_4\), represents the actual number and types of atoms in a molecule. In a formula, the subscript numbers indicate how many of each atom are in the molecule, and if there's no number, it means there's just one atom of that kind.

From \(H_2SO_4\)'s chemical formula, we know it's composed of 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms. This information is crucial for calculating molar masses and carrying out stoichiometric conversions, like determining the mass of sulfur in a given mass of \(H_2SO_4\), which ties back to understanding the stoichiometry of the compound.

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Most popular questions from this chapter

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives \(3.38 \mathrm{~g}\) carbon dioxide, \(0.690 \mathrm{~g}\) of water and no other products. A volume of \(10.0\) litre (Measured at STP) of this welding gas is found to weigh \(11.6 \mathrm{~g}\). Calculate : (i) empirical formula, (ii) molar mass of the gas, and (iii) moleculat formula.

\(30 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{BaCl}_{2}\) is mixed with \(40 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). What is the weight of \(\mathrm{BaSO}_{4}\) formed? $$ \mathrm{BaCl}_{2}+\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \longrightarrow \mathrm{BaSO}_{4}+\mathrm{AlCl}_{3} $$

Calculate the amount of oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)\) required to obtain \(250 \mathrm{~mL}\) of deci-molar solution.

Calculate normality of \(\mathrm{NH}_{4} \mathrm{OH}\) when \(2 \mathrm{~g}\) is present in \(800 \mathrm{~mL}\) solution. Also calculate its molarity.

(a) When \(4.2 \mathrm{~g} \mathrm{NaHCO}_{3}\) is added to a solution \(\mathrm{of} \mathrm{CH}_{3} \mathrm{COOH}\) weighing \(10.0 \mathrm{~g}\), it is observed that \(2.2 \mathrm{~g} \mathrm{CO}_{2}\) is released into atmosphere. The residue is found to weigh \(12.0 \mathrm{~g}\). Show that these observations are in agreement with the low of conservation of weigh. (b) If \(6.3 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}\) are added to \(15.0 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}\) solution. The residue is found to weigh \(18.0 \mathrm{~g}\). What is the mass of \(\mathrm{CO}_{2}\) released in this reaction.

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