Question

How many grams of calcium nitrate (Ca(NO \(_{3} )_{2} )\) would you need to prepare 3.00 L of a 0.500\(M\) solution?

Short answer

To prepare 3.00 L of a 0.500 M solution of calcium nitrate (Ca(NO\(_3\))\(_2\)), you would need 245 grams of calcium nitrate.

Step-by-step solution

Identify the given values

In this problem, we are given: - Molarity (M) = 0.500 M - Volume (V) = 3.00 L - Molar mass of calcium nitrate (Ca(NO3)₂) = 164.09 g/mol

Use the formula to calculate the mass of calcium nitrate needed

Using the formula mass = molarity × volume × molar mass of solute, we can calculate the mass of calcium nitrate needed. Mass = M × V × Molar mass of Ca(NO3)₂ Plug in the given values: Mass = (0.500 mol/L) × (3.00 L) × (164.09 g/mol)

Calculate the mass of calcium nitrate

Multiply the values: Mass = (0.500) × (3.00) × (164.09) Mass = 245.135 g

Round the answer to an appropriate number of significant figures

Since we have 3 significant figures in the given volume and molarity, we should round off our final answer to 3 significant figures as well: Mass = 245 g Thus, you would need 245 grams of calcium nitrate (Ca(NO3)₂) to prepare 3.00 L of a 0.500 M solution.

Key concepts

Molarity Calculations

Understanding how to calculate molarity is foundational in chemistry, particularly when preparing solutions. Molarity, symbolized by the capital letter M, is a measure of concentration that expresses the moles of solute per liters of solution. Its general formula is: \[\begin{equation} Molarity (M) = \frac{moles\ of\ solute}{liters\ of\ solution} \end{equation}\]To find the number of moles needed, you can rearrange the formula:\[\begin{equation} moles = Molarity\ (M) \times Volume\ (L) \end{equation}\]However, in practical scenarios like the given exercise, the desired quantity is often the mass of a substance, not moles. Here, knowing the molar mass of the solute (in grams per mole) allows us to convert moles to grams using the formula:\[\begin{equation} Mass = Molarity\ (M) \times Volume\ (L) \times Molar\ mass\ (g/mol) \end{equation}\]This becomes especially useful in laboratory settings for creating specific solution concentrations by weighing out amounts of a substance. The ease of creating accurate molar solutions by careful calculation benefits various applications from research to pharmaceutical preparations.

When utilizing molarity calculations, accuracy is key. While it may seem simple to plug values into a formula, understanding the relationship between moles, volume, and molar mass grounds you in the science of solution preparation.

Stoichiometry

Stoichiometry is the set of mathematical relationships between the amounts of reactants and products in a chemical reaction. It is derived from the Greek words 'stoicheion' (element) and 'metron' (measure). In the context of molarity and solution preparation, stoichiometry comes into play when considering reaction equations and the subsequent formation of products from reactants in a solution.

For instance, if you are reacting a substance with calcium nitrate in a solution, stoichiometry would tell you how much of each reactant you need to produce a desired amount of product. It also can dictate the ratios when preparing solutions for titrations or when a particular reaction's completion is desired. The stoichiometric coefficients from balanced chemical equations guide the conversion between moles of different substances.

Applying Stoichiometry to Molarity

In the exercise provided, stoichiometry is implicitly used when calculating the amount of calcium nitrate required to achieve the desired molarity. While the exercise does not involve a reaction, the principles of stoichiometric calculations are inherently present in the conversion of molar mass to grams, ensuring the right mass of solute for the correct volume of solvent. Accurate stoichiometric calculations are essential for creating precise chemical solutions and for scaling up reactions for industrial applications or scaling down for microscale experiments.

Significant Figures

When it comes to reporting measurements in chemistry, significant figures (sig figs) are pivotal. They convey the precision of the measurement tools used and the inherent uncertainty in the measurements. Every digit in a significant figure is meaningful, helping to ensure the accuracy of scientific communication.

Sig figs are determined by the equipment or method used to take a measurement. For instance, if a volume is measured with a graduated cylinder that reads up to 0.01 liters, then a volume measurement of 3.00 L has three significant figures. The zeroes indicate the precision of the measurement, not just placeholders.

In calculations like the one in this exercise, sig figs guide the final answer's precision. After performing the multiplication, the result must be rounded to the correct number of sig figs to reflect the precision of the given information. As in the provided solution, even though the unrounded answer was 245.135 g, it was rounded to 245 g to match the three significant figures from the initial given volume and molarity.

Why Sig Figs Matter

Keeping track of significant figures is crucial in the real world where the accuracy of measurement can impact outcomes significantly. In pharmaceutical preparations, for example, the correct dosage of a drug depends on accurate measurements. In engineering, structural integrity can hinge on precise calculations. Accuracy and clear communication in measurements ensure reproducibility and reliability in scientific practice.