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How many molecules of water are there in \(10.0 \mathrm{~g}\) of water?

Short Answer

Expert verified
There are approximately \(3.34 \times 10^{23}\) molecules of water in 10.0 g of water.

Step by step solution

01

Determine the molar mass of water (H鈧侽)

First, find the molar mass of water (H鈧侽). This is done by adding the molar masses of the individual elements present in one molecule of water: Molar mass of H鈧侽 = (2 脳 molar mass of H) + molar mass of O The molar mass of hydrogen is 鈧丠 鈮 1.008 g/mol and the molar mass of oxygen is 鈧佲倖O 鈮 16.00 g/mol. Molar mass of H鈧侽 = (2 脳 1.008 g/mol) + 16.00 g/mol = 18.016 g/mol
02

Convert the mass of water (10.0 g) into moles using the molar mass

Now we need to convert 10.0 g of water into moles using the molar mass of water we found in step 1. Moles = mass / molar mass Moles = (10.0 g) / (18.016 g/mol) = 0.555 moles
03

Multiply the number of moles by Avogadro's number to find the number of water molecules

Finally, multiply the number of moles by Avogadro's number (6.022 脳 10虏鲁 molecules/mol) to find the number of water molecules: Number of water molecules = (number of moles) 脳 Avogadro's number Number of water molecules = (0.555 moles) 脳 (6.022 脳 10虏鲁 molecules/mol) = 3.34 脳 10虏鲁 molecules Therefore, there are about 3.34 脳 10虏鲁 molecules of water in 10.0 g of water.

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Key Concepts

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

Molar Mass Calculation
Understanding the molar mass of a substance is critical for many areas of chemistry, particularly when working with chemical equations and reactions. The molar mass tells us how much one mole of a substance weighs. To calculate it, we simply sum up the atomic masses of each element in a molecule, which are found on the periodic table, multiplied by the number of atoms of that element in the molecule.

For instance, in the exercise given, we calculated the molar mass of water (H鈧侽) by adding together the atomic masses of two hydrogens (approximately 1.008 g/mol each) and one oxygen (approximately 16.00 g/mol). This calculation resulted in the molar mass of water as 18.016 g/mol. It's essential to ensure that the calculation is based on the correct number of atoms of each element in the molecule for an accurate molar mass.
Avogadro's Number
Avogadro's number is a cornerstone in stoichiometry and provides a bridge between the atomic scale and the macroscopic world. It is defined as the number of units (usually atoms or molecules) in one mole of any substance and is approximately equal to 6.022 脳 10虏鲁.

This concept allows chemists to count particles by weighing, as we often deal with quantities too large to count individually. When we refer to Avogadro's number, we're talking about a scale factor that tells us how many particles there are in a mole of a substance. In our exercise, it helps us understand that a mole represents a set number of molecules, which in this case, helps us calculate the huge number of water molecules in a seemingly small quantity of water.
Stoichiometry
Stoichiometry can be daunting, but it's essentially just the math behind chemistry, dealing with the quantitative relationships between reactants and products in a chemical reaction. In practice, it involves calculations using molar masses, Avogadro's number, and balance equations to predict the amounts of products and reactants.

In the exercise, we practiced stoichiometry by first calculating the molar mass, then using that figure to convert grams to moles, and finally, applying Avogadro's number to find the total molecules. This is a standard stoichiometric process, allowing us to move from the mass of a substance to an actual particle count, demonstrating the practical application of stoichiometry in solving chemistry problems.
Chemical Composition
The chemical composition of a substance refers to the identity and ratio of the elements that make it up. Every molecule has a unique composition that defines its properties and behaviors. It is often represented by a chemical formula, such as H鈧侽 for water, which shows that each molecule is comprised of two hydrogen atoms bonded to one oxygen atom.

Understanding a substance's chemical composition is crucial for all chemical calculations, including those for molar mass, stoichiometric conversions, and molecular counts illustrated in our exercise. By knowing the composition, we can discern the proper ratios of elements and predict the outcomes of chemical reactions, allowing us to delve into the molecular world and make sense of matter's intricate details.

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Most popular questions from this chapter

Consider the balanced chemical equation $$ \begin{aligned} &\mathrm{Fe}(\mathrm{CO})_{5}(s)+2 \mathrm{PF}_{3}(l)+\mathrm{H}_{2}(g) \rightarrow \\ &\mathrm{Fe}(\mathrm{CO})_{2}\left(\mathrm{PF}_{3}\right)_{2} \mathrm{H}_{2}(s)+3 \mathrm{CO}(g) \end{aligned} $$ (a) How many grams of \(\mathrm{CO}\) could be produced from \(10.0 \mathrm{~g}\) of \(\mathrm{PF}_{3}\), excess \(\mathrm{Fe}(\mathrm{CO})_{5}\), and excess \(\mathrm{H}_{2} ?\) (b) How many grams of \(\mathrm{CO}\) could be produced from \(5.0\) moles of \(\mathrm{Fe}(\mathrm{CO})_{5}, 8.0\) moles of \(\mathrm{PF}_{3}\), and \(6.0\) moles of \(\mathrm{H}_{2}\) ? (c) How many moles of \(\mathrm{CO}\) could be produced from \(25.0 \mathrm{~g}\) of \(\mathrm{Fe}(\mathrm{CO})_{5}, 10.0 \mathrm{~g}\) of \(\mathrm{PF}_{3}\), and excess \(\mathrm{H}_{2} ?\) (d) The density of hydrogen gas at room temperature and atmospheric pressure is \(0.0820 \mathrm{~g} / \mathrm{L}\). When \(5.00 \mathrm{~L}\) of hydrogen gas at room temperature and atmospheric pressure is mixed with excess \(\mathrm{Fe}(\mathrm{CO})_{5}\) and excess \(\mathrm{PF}_{3}\), the mass of \(\mathrm{CO}\) collected is \(13.5 \mathrm{~g}\). What is the theoretical yield (in grams) and the percent yield of \(\mathrm{CO}\) ?

For the reaction \(\mathrm{SiCl}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{SiO}_{2}+\mathrm{HCl}\), (a) Balance the equation. (b) If you want to make \(120.0 \mathrm{~g}\) of \(\mathrm{HCl}\), how many grams of \(\mathrm{SiCl}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) do you need? (c) How many grams of \(\mathrm{SiO}_{2}\) are produced from the quantities calculated in (b)?

A \(1.000-g\) sample of a liquid that contains only carbon and hydrogen burns in oxygen to produce \(1.284 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) (a) What are the mass percents of the elements present in this sample? (b) What is the empirical formula for this compound? (c) The molar mass of this compound is determined to be about \(71 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound? (Hint: When attempting this problem, understand that all of the carbon in the compound burned ends up as \(\mathrm{CO}_{2}\), and all of the hydrogen in the compound burned ends up as \(\mathrm{H}_{2} \mathrm{O}\). Also, there is only 1 mole of \(C\) per mole of \(\mathrm{CO}_{2}\), but there are 2 moles of \(\mathrm{H}\) per mole of \(\mathrm{H}_{2} \mathrm{O} .\) )

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?

A compound used as an insecticide that contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{Cl}\) is subjected to combustion analysis, yielding \(55.55 \% \mathrm{C}\) and \(3.15 \% \mathrm{H}\). (a) What is the empirical formula of this compound? (b) What is the molecular formula if its empirical formula is \(1 / 3\) the mass of its actual formula?

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