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Chapter 9: Problem 71

Suppose you wanted 1 billion \(\left(1.00 \times 10^{9}\right)\) water molecules and you didn't have time to sit and count them out. How many grams of water would you need to get 1 billion water molecules?

Short Answer

Expert verified
To obtain 1 billion ($1.00 \times 10^9$) water molecules, you would need approximately \(2.99 \times 10^{-14}\) grams of water.

Step by step solution

01

Determine the number of water molecules per mole

We know that one mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles. In this case, the particles are water molecules, so one mole of water contains \(6.022 \times 10^{23}\) water molecules.
02

Calculate the molar mass of water

Water is composed of two hydrogen atoms and one oxygen atom, so its chemical formula is H₂O. We need to determine the molar mass of water, which is the sum of the molar masses of all the atoms in the molecule. The molar mass of hydrogen is approximately 1 g/mol and the molar mass of oxygen is approximately 16 g/mol. Therefore, the molar mass of water is approximately \(2 \times 1 + 16 = 18\text{ g/mol}\).
03

Calculate the number of moles required

We want 1 billion water molecules \(\left(1.00 \times 10^9\right)\). To find out how many moles this is, we can divide the number of desired water molecules by Avogadro's number: \[ \text{Number of moles} = \frac{1.00 \times 10^9 \text{ water molecules}}{6.022 \times 10^{23} \text{ water molecules/mol}} \]
04

Calculate the mass of water needed

To find the mass of water needed, we can multiply the number of moles required (calculated in Step 3) by the molar mass of water: \[ \text{Mass of water} = (\text{Number of moles}) \times (18\text{ g/mol}) \] Now let's perform the calculations:
05

Perform the calculations

\[ \text{Number of moles} = \frac{1.00 \times 10^9 \text{ water molecules}}{6.022 \times 10^{23} \text{ water molecules/mol}} \approx 1.66 \times 10^{-15} \text{ moles} \] \[ \text{Mass of water} = (1.66 \times 10^{-15} \text{ moles}) \times (18\text{ g/mol}) \approx 2.99 \times 10^{-14} \text{ g} \] So, you would need approximately \(2.99 \times 10^{-14}\) grams of water to have 1 billion water molecules.

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Most popular questions from this chapter

A compound is \(91.77 \%\) by mass \(\mathrm{Si}\) and \(8.23 \%\) by mass \(\mathrm{H}\) and has a molar mass of approximately \(122 \mathrm{~g} / \mathrm{mol}\). What is its molecular formula?

Determine the empirical formula of the compound with the following mass percents of the elements present: \(58.5 \% \mathrm{C} ; 4.91 \% \mathrm{H} ; 19.5 \% \mathrm{O} ; 17.1 \% \mathrm{~N}\).

A \(2.230-\mathrm{g}\) sample of a solid is subjected to combustion analysis, yielding \(76.59 \% \mathrm{C}\) and \(6.39 \% \mathrm{H}\). It may also contain oxygen. (a) What is the empirical formula for this compound? (b) The molar mass of this compound is determined to be about \(94 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

(a) What is the molar mass of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) ?\) (b) What is the mass of \(1.25\) moles of sucrose? (c) How many sucrose molecules are in \(1.25\) moles? (d) How many hydrogen atoms are in \(1.25\) moles of sucrose? (e) What is the mass in grams of the hydrogen atoms in part (d)?

Consider the following decomposition reaction in which \(47.20 \mathrm{~g}\) of some compound is decomposed into its elements.
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