Question

You go to a convenience store to buy candy and find the owner to be rather odd. He allows you to buy pieces only in multiples of four, and to buy four, you need \(\$ 0.23 .\) He allows you only to use 3 pennies and 2 dimes. You have a bunch of pennies and dimes, and instead of counting them, you decide to weigh them. You have \(636.3 \mathrm{g}\) of pennies, and each penny weighs an average of \(3.03 \mathrm{g}\). Each dime weighs an average of \(2.29 \mathrm{g}\). Each piece of candy weighs an average of \(10.23 \mathrm{g}\) a. How many pennies do you have? b. How many dimes do you need to buy as much candy as possible? c. How much would all of your dimes weigh? d. How many pieces of candy could you buy (based on the number of dimes from part b)? e. How much would this candy weigh? f. How many pieces of candy could you buy with twice as many dimes?

Short answer

a. The number of pennies is \(p = \frac{636.3}{3.03} \approx 210\). b. The number of dimes needed is \(d = \frac{2}{3}p \approx 140\). c. The total weight of dimes is \(w_d = d \times 2.29 \approx 320.6 \mathrm{g}\). d. The number of candies you could buy is \(c = \frac{d}{2} \approx 70\). e. The total weight of candies would be \(w_c = c \times 10.23 \approx 716.1 \mathrm{g}\). f. The number of candies you could buy with twice as many dimes is \(c_2 = \frac{2d}{2} = d \approx 140\).

Step-by-step solution

Determine the number of pennies

To find out how many pennies we have, we will divide the total weight of the pennies (636.3 g) by the average weight of each penny (3.03 g). Let \(p\) denote the number of pennies: \(p = \frac{636.3}{3.03}\)

Calculate how many dimes are needed

Since we can only buy candy in multiples of four, we need to find out how many dimes we need to buy the maximum number of candies using only 3 pennies and 2 dimes. We will use the ratio of three pennies and two dimes and the number of pennies we have in Step 1 to find the total number of dimes needed. Let \(d\) be the number of dimes needed: \(d = \frac{2}{3}p\)

Measure the total weight of dimes

We can find the total weight of the dimes we have by multiplying the number of dimes from Step 2 by the average weight of a dime (2.29 g). \(w_d = d \times 2.29\)

Calculate the number of candies

From earlier, we know that we need 3 pennies and 2 dimes per piece of candy. We can calculate the number of candies we can buy based on the number of dimes we have from Step 2. Let \(c\) be the number of candies we can buy: \(c = \frac{d}{2}\)

Calculate the total weight of candies

To find the total weight of the candies, we will multiply the number of candies we can buy (from Step 4) by the average weight of a piece of candy (10.23 g). \(w_c = c \times 10.23\)

Calculate the number of candies with twice as many dimes

Finally, we want to find out how many pieces of candy we could buy if we had twice as many dimes. We can calculate this by doubling the number of dimes in our possession. Let \(c_2\) be the number of candies we can buy with twice as many dimes: \(c_2 = \frac{2d}{2}\)

Key concepts

Stoichiometry in Everyday Transactions

Stoichiometry is a core concept in chemistry that relates to the quantitative relationships between reactants and products in a chemical reaction. However, it can also be applied to everyday scenarios such as the one in this candy-buying exercise. In our case, stoichiometry is utilized to analyze the exchange between money—in the form of pennies and dimes—and candy pieces. This is analogous to a chemical reaction where reactants are converted into products in certain ratios.

By understanding the 'reaction' ratio of 3 pennies and 2 dimes for every 4 pieces of candy, we can use stoichiometric calculations to determine the maximum amount of candy that can be bought. It's crucial to note how ratios play a significant role in these calculations, ensuring that the exchange follows the predetermined rules, much like ensuring that a chemical equation is balanced before calculating the amount of products formed from given reactants.

Unit Conversion: From Grams to Candy

Unit conversion is essential in chemistry problem-solving because it allows us to change different units of measure such that we can compare, combine, or analyze different types of data. In our exercise, we had to convert the weight of coins into the number of coins by using the average weight of each coin type, and then further translate that into the amount of candy that could be purchased.

For instance, calculating the number of pennies involves dividing the total weight by the average weight per penny, i.e.,
\( p = \frac{636.3 \text{g}}{3.03 \text{g/penny}} \). The division here is a unit conversion, changing grams into 'penny count'. Similarly, for dimes, we had to consider the weight of each dime to determine the total weight that a certain number of dimes would have. That's precisely why unit conversions are the subtle heroes of chemistry; they connect different kinds of measurements in a meaningful way.

The Mole Concept Extending Beyond Chemical Substances

The mole concept, typically used to count atoms and molecules in chemistry, is a way to deal with large quantities of very small entities by grouping them into 'moles'. One mole represents Avogadro's number (\(6.022 \times 10^{23}\)) of items, whether atoms, ions, or in this case, coins.

Imagine if we used a unit equivalent to 'a mole of coins' for everyday transactions! It would be impratical due to the enormity of the number. However, similar to how we use moles to simplify calculations in chemistry, in this exercise, we group coins into manageable units (pennies in sets of 3, and dimes in sets of 2) to simplify the transaction process. This grouping concept is core to understanding stoichiometry as it facilitates the handling of large numbers in a systematized and calculable manner. In the realm of education, especially chemistry, breaking down the intimidating 'mole' into familiar, relatable concepts like this can greatly aid in student comprehension.