- Define the given mass as \(A = 6.37 \text{ grams}\).
- Use the molar mass of NaCl, \(58.5 \text{ g/mol}\), as the conversion factor \(\frac{B}{C} = \frac{1 \text{ mole}}{58.5 \text{ grams}}\).
- The equation becomes \(6.37 \times \frac{1}{58.5} = \text{moles NaCl}\).
- Therefore: \(A = \boxed{6.37}\), \(B = \boxed{1}\), \(C = \boxed{58.5}\).
Explanation
- Analysis of the Conversion Problem
To determine the molar amount of table salt (NaCl) present in a given mass, we use the molecular weight as a conversion factor. We are given the following information:
- The mass of the salt is \(6.37 \text{ grams}\).
- The molecular weight of NaCl is \(58.5 \text{ g/mol}\), which means that 1 mole of NaCl weighs \(58.5 \text{ grams}\).
- The equation to solve is \(A \times \frac{B}{C} = \text{moles NaCl}\).
Dimensional analysis requires that the units cancel out correctly so that we are left with the target unit, which is "moles". We start with a value in grams (A), multiply by a fraction where the unit of "grams" is in the denominator (C) to cancel out the initial "grams", and the unit of "moles" is in the numerator (B).
- Setting Up the Dimensional Analysis
We want to convert the mass of salt, \(6.37 \text{ grams}\), into moles using the conversion factor derived from the molar mass.
- Assign the known quantity, \(6.37 \text{ grams}\), to \(A\).
- To convert from grams to moles, we divide by the molar mass (\(58.5 \text{ g/mol}\)). This is equivalent to multiplying by \(\frac{1 \text{ mole}}{58.5 \text{ grams}}\).
- Comparing this to the formula \(A \times \frac{B}{C}\), we assign \(B = 1 \text{ mole}\) and \(C = 58.5 \text{ grams}\).
- The setup becomes:
\[(6.37 \text{ g}) \times \frac{1 \text{ mole}}{58.5 \text{ g}} = \text{moles NaCl}\]
All units cancel out correctly (\(\frac{\text{g} \times \text{mol}}{\text{g}} = \text{mol}\)), leaving us with the desired unit.
- Final Calculation and Results
Following the established unit conversion, the values for the variables are:
- \(A = 6.37 \text{ grams}\)
- \(B = 1 \text{ mole}\)
- \(C = 58.5 \text{ grams}\)
Performing the calculation:
\[\frac{6.37}{58.5} \approx 0.108888... \text{ moles}\]
Rounding to significant figures as appropriate for the input, we find the amount of salt to be approximately \(0.109 \text{ moles}\). Thus, the required values are:
\(A = \boxed{6.37}\), \(B = \boxed{1}\), \(C = \boxed{58.5}\).
Examples
Dimensional analysis is a fundamental skill in chemistry and physics for converting between units, such as calculating the number of moles of a substance from its mass, or determining travel time based on speed and distance. For example, if you know a car travels at a speed of 60 miles per hour and you want to know how far it travels in 3 hours, you set up the equation: \((3 \text{ hours}) \times \frac{60 \text{ miles}}{1 \text{ hour}} = 180 \text{ miles}\). This method ensures that units match and mistakes are minimized in complex calculations.