Question

In this experiment it takes about 6 microliters of solution to produce a spot \(8 \mathrm{mm}\) in diameter. If the \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) solution contains about \(6 \mathrm{g} \mathrm{Cu}^{2+}\) per liter, how many micrograms of \(\mathrm{Cu}^{2+}\) ion are there in one spot?

Short answer

The amount of Cu²⁺ ions in one spot can be calculated by first converting the concentration to g/µL, then finding the mass of Cu²⁺ ions in the given volume, and finally converting the mass to micrograms. By performing these calculations, we get: \( \left(\frac{6 \; g/L}{1000000 \;µL / L} × 6\;µL\right) × 1000000\;µg/g \), which gives the amount of Cu²⁺ ions in one spot in micrograms.

Step-by-step solution

Convert concentration from grams per liter to grams per microliter

We are given that the concentration of Cu²⁺ ions is 6 g/L. We need to change this unit to g/µL. 1 L = 1000000 µL, therefore, to convert the concentration: \( Concentration \; (g/µL) = \frac{6 \;g/L}{1000000 \;µL/L} \)

Calculate the mass of Cu²⁺ ions in the given volume

Now that we know the concentration of Cu²⁺ ions in g/µL, we can find the mass of Cu²⁺ ions in one spot by using the given volume, 6 µL. Mass = Concentration × Volume \( Mass = \frac{6 \;g/L}{1000000 \;µL/L} × 6 \;µL \)

Convert mass to micrograms

The mass we calculated in the previous step is in grams. To change it to micrograms, we need to multiply by 1000000. Micrograms = Mass × 1000000 \( Micrograms = \left(\frac{6 \;g/L}{1000000 \; µL / L} × 6\; µL\right) × 1000000\;µg/g \) Now, calculating the value, we get the amount of Cu²⁺ ions in one spot in micrograms.

Key concepts

Molarity and Concentration

Understanding molarity and concentration is a fundamental concept in stoichiometry, central to comprehending the composition of solutions in chemistry. Molarity, designated by the symbol 'M,' is a measure of how many moles of solute are present in a liter of solution. Moles represent a standard number of particles (\( 6.022 \times 10^{23} \) particles), named Avogadro's number, which provides a consistent way to describe quantities of substances.

When a problem states that a \( \mathrm{Cu}^{2+} \) solution contains roughly 6 grams of \( \mathrm{Cu}^{2+} \) ions per liter, it's referring to the solution's concentration. This concentration tells us the amount of solute in a given volume of solution. To solve stoichiometry problems, we convert this concentration into a more usable form, depending on the volume we’re working with, which leads to our next step: unit conversion.

Unit Conversion

A crucial step in stoichiometry, and science in general, is unit conversion. This process allows us to convert measurements from one unit to another, enabling calculations when units differ. Often, this is done using conversion factors, which are ratios that express how a quantity in one unit is equal to the quantity in another.

For instance, in our exercise, we need to convert the concentration of \( \mathrm{Cu}^{2+} \) from grams per liter (g/L) to grams per microliter (g/µL). The conversion factor in this case is the relationship between liters and microliters. Since 1 liter equals 1,000,000 microliters, we divide the concentration in grams per liter by 1,000,000 to find grams per microliter. This step makes the calculation for a smaller volume, such as the one needed for the spot in our experiment, straightforward.

Calculating Mass from Volume and Concentration

After understanding molarity and mastering unit conversion, it's time to calculate the mass of a solute in a solution given its volume and concentration. The formula for this calculation is simple: Mass = Concentration × Volume.

In the problem presented, we are given the volume of the spot produced by the solution, which is 6 microliters. Using the volume and the converted concentration, the mass in grams is easily calculated. Since the exercise requires the mass in micrograms, a final conversion is necessary. By multiplying the mass in grams by 1,000,000 (the number of micrograms in a gram), we obtain the final result, leading to a complete understanding of the amount of \( \mathrm{Cu}^{2+} \) ions in one spot of our experiment.