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Chapter 1: Problem 67

A sample of ascorbic acid (vitamin \(\mathrm{C} )\) is synthesized in the laboratory. It contains 1.50 \(\mathrm{g}\) of carbon and 2.00 \(\mathrm{g}\) of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains 6.35 \(\mathrm{g}\) of carbon. According to the law of constant composition, how many grams of oxygen does it contain?

Short Answer

Expert verified
The mass ratio of carbon to oxygen in the first sample of ascorbic acid is \(0.75\). Using this ratio, there are approximately \(8.47 \mathrm{g}\) of oxygen in the second sample, which contains \(6.35 \mathrm{g}\) of carbon.

Step by step solution

01

Calculate the mass ratio of the elements in the first sample

First, we need to find out the mass ratio of carbon and oxygen in the first sample. We are given 1.50 grams of carbon and 2.00 grams of oxygen, so the mass ratio of carbon to oxygen is: \[\frac{\text{mass of carbon}}{\text{mass of oxygen}} = \frac{1.50}{2.00} = 0.75\]
02

Use the mass ratio to find the mass of oxygen in the second sample

Now, we'll use the calculated mass ratio from step 1 to find the mass of oxygen in the second sample. We're given that the second sample contains 6.35 grams of carbon. Let 'x' be the mass of oxygen in the second sample. \[0.75 = \frac{6.35}{x}\]
03

Solve for x

Now, we need to solve for 'x' to find out the mass of oxygen in the second sample: \[x = \frac{6.35}{0.75}\] \[x \approx 8.47\] So, there are approximately 8.47 grams of oxygen in the second sample of ascorbic acid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Ascorbic Acid
Ascorbic acid, also known as vitamin C, is an essential nutrient for the human body. It's vital for the growth and repair of tissues. The chemical formula for ascorbic acid is \( \text{C}_6\text{H}_8\text{O}_6 \). This signifies six carbon atoms, eight hydrogen atoms, and six oxygen atoms in each molecule.

Ascorbic acid is found naturally in citrus fruits, such as oranges and lemons. But it can also be synthesized in the laboratory for various uses, such as in supplements and fortified foods. Understanding its composition is crucial in fields like nutrition and chemistry, as its structure and elemental makeup influence its function and benefits. By examining its components, we can see the intricate balance of elements that contribute to its effectiveness as a nutrient.
Concept of Mass Ratio
The mass ratio of a substance refers to the proportion of elements by mass within a compound. It's a crucial concept in chemistry that follows the law of constant composition, stating that a given compound will always contain the same proportion of elements by mass.

For instance, in the original exercise, the given ascorbic acid sample has 1.50 grams of carbon and 2.00 grams of oxygen. Therefore, its mass ratio is calculated as:
  • Mass ratio = \( \frac{1.50}{2.00} = 0.75 \)
This ratio remains consistent across any sample of the same compound, whether it's synthesized in a lab or extracted from fruits.

Using the mass ratio, one can predict the mass of another element in a different sample of the same compound. This can be particularly useful in determining the constituency of larger samples, and ensures the consistency of chemical analysis across different environments and sources.
Chemical Sample Analysis in Practice
Chemical sample analysis involves examining the composition of substances to determine their structure and elemental makeup. This process is vital in achieving accurate results and understanding substances on a molecular level.

To analyze a chemical sample effectively:
  • Identify the known quantities or masses of elements in a sample.
  • Apply the concept of mass ratios to predict unknown masses.
  • Utilize mathematical equations to calculate these unknowns accurately.
In the given exercise, after calculating the mass ratio of carbon to oxygen in the first sample of ascorbic acid (0.75), we used it to find the mass of oxygen in the second sample. By setting up the equation \( 0.75 = \frac{6.35}{x} \) and solving for \( x \), the mass of oxygen was determined to be approximately 8.47 grams.

This methodical approach underscores the importance of following consistent rules of chemical composition and analysis, allowing for reliable and repeatable results regardless of the source or size of the chemical sample.

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Most popular questions from this chapter

The distance from Earth to the Moon is approximately \(240,000 \mathrm{mi.}\) (a) What is this distance in meters? (b) The peregrine falcon has been measured as traveling up to 350 \(\mathrm{km} /\) hr in a dive. If this falcon could fly to the Moon at this speed, how many seconds would it take? (c) The speed of light is \(3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) How long does it take for light to travel from Earth to the Moon and back again? (\boldsymbol{d} ) ~ E a r t h ~ t r a v e l s ~ a r o u n d ~the Sun at an average speed of 29.783 \(\mathrm{km} / \mathrm{s} .\) Convert this speed to miles per hour.

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A \(25.0-\) mL portion of the liquid had a mass of 21.95 \(\mathrm{g} .\) A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as 0.8787 \(\mathrm{g} / \mathrm{mL} .\) Is the calculated density of benzene at \(15^{\circ} \mathrm{C}\) as 0.8787 \(\mathrm{g} / \mathrm{mL} .\) Is the calculated density in agreement with the tabulated value? (b) An experiment requires 15.0 \(\mathrm{g}\) of cyclohexane, whose density at \(25^{\circ} \mathrm{C}\) is 0.7781 \(\mathrm{g} / \mathrm{mL}\) . What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of 5.0 \(\mathrm{cm} .\) What is the mass of the sphere if lead has a density of 11.34 \(\mathrm{g} / \mathrm{cm}^{3} ?\) (The volume of a sphere is \((4 / 3) \pi r^{3},\) where \(r\) is the radius.)

(a) To identify a liquid substance, a student determined its density. Using a graduated cylinder, she measured out a 45 -mL. sample of the substance. She then measured the mass of the sample, finding that it weighed 38.5 \(\mathrm{g}\) . She knew that the substance had to be either isopropylalcohol (density 0.785 \(\mathrm{g} / \mathrm{mL}\) )or toluene (density 0.866 \(\mathrm{g} / \mathrm{mL} ) .\) What are the calculated density and the probable identity of the substance? (b) An experiment requires 45.0 \(\mathrm{g}\) of ethylene glycol, a liquid whose density is 1.114 \(\mathrm{g} / \mathrm{mL}\) . Rather than weigh the sample on a balance, a chemist chooses to dispense the liquid using a graduated cylin-der. What volume of the liquid should he use? (c) Is a graduated cylinder such as that shown in Figure 1.21 likely to afford the accuracy of measurement needed? (d) A cubic piece of metal measures 5.00 \(\mathrm{cm}\) on each edge. If the metal is nickel, whose density is \(8.90 \mathrm{g} / \mathrm{cm}^{3},\) what is the mass of the cube?

Using your knowledge of metric units, English units, and the information on the back inside cover, write down the con- version factors needed to convert (a) mm to nm, (b) mg to kg, (c) km to ft, (d) in. \(^{3}\) to \(\mathrm{cm}^{3} .\)

(a) The temperature on a warm summer day is \(87^{\circ} \mathrm{F}\) . What is the temperature in \(^{\circ} \mathrm{C} ?\) (b) Many scientific data are reported at \(25^{\circ} \mathrm{C}\) . What is this temperature in kelvins and in degrees Fahrenheit? (c) Suppose that a recipe calls for an oven temperature of \(400^{\circ} \mathrm{F}\) . Convert this temperature to degrees Celsius and to kelvins. (d) Liquid nitrogen boils at 77 \(\mathrm{K}\) . Convert this temperature to degrees Fahrenheit and to degrees Celsius.
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