Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 8: Problem 30

The water content of an \(875.4-\mathrm{mg}\) sample of cheese is determined with a moisture analyzer. What is the \(\% \mathrm{w} / \mathrm{w} \mathrm{H}_{2} \mathrm{O}\) in the cheese if the final mass was found to be \(545.8 \mathrm{mg} ?\)

Short Answer

Expert verified
The water content in the cheese is approximately 37.65\% w/w.

Step by step solution

01

Calculate the Mass of Water Lost

To find the mass of water lost, subtract the final mass of the cheese from the initial mass. The initial mass of the cheese is given as 875.4 mg, and the final mass (after water loss) is 545.8 mg. Therefore, the mass of the water lost is: \[ \text{Mass of } \mathrm{H}_2\mathrm{O} = 875.4 \mathrm{mg} - 545.8 \mathrm{mg} = 329.6 \mathrm{mg} \]
02

Calculate the Water Content as a Percent by Weight

To find the percentage by weight of water in the cheese, divide the mass of water by the initial mass of cheese and multiply by 100. Using the previously calculated mass of water: \[ \% \mathrm{w} / \mathrm{w} \mathrm{H}_2\mathrm{O} = \left(\frac{329.6 \mathrm{mg}}{875.4 \mathrm{mg}}\right) \times 100 \approx 37.65\% \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Moisture Analysis
Moisture analysis is a critical component in the field of analytical chemistry, often employed in the food industry, pharmaceuticals, and many other sectors. It involves determining the amount of water present in a material, which is essential for quality control and shelf life predictions.
A moisture analyzer is a tool specifically designed for this purpose. It accurately measures the weight of a sample before and after drying to assess water loss. Utilizing these measurements, one can determine the moisture content.
  • These devices use heat to evaporate moisture from the sample.
  • They provide direct moisture readings, saving time and effort compared to classic methods.
  • They help in maintaining product consistency and quality across batches.
Accurate moisture analysis helps in understanding the product better and ensures its stability, effectiveness, and quality during storage and usage.
Mass Calculation
When performing moisture analysis, it's vital to accurately calculate the mass of the substance both before and after drying. Mass calculation helps keep track of the exact amount of water that evaporates during the process.
To find the change in mass, simple arithmetic is used. For example, subtracting the final mass of the cheese after it has dried from its initial mass gives us the mass of the moisture lost. The process is straightforward but requires precision to ensure accurate results:
  • Initial mass is measured before drying.
  • Final mass is determined after the moisture is removed.
  • The difference gives the mass of the evaporated water.
Proper mass calculation is crucial as any errors can lead to significant deviations in the estimated water content.
Percentage Composition
Percentage composition is a way of expressing the concentration of a component in a mixture. In this context, it refers to the percentage of water in the cheese by weight. Calculating this percentage is essential for understanding the moisture content in the context of the entire sample.
Calculating percentage composition involves a simple formula: taking the mass of the component of interest and dividing it by the total mass of the mixture, then multiplying by 100 to convert it into a percentage.

Formula

The formula to calculate the percentage of water is:\[\% \text{ w/w } \mathrm{H}_2\mathrm{O} = \left(\frac{\text{Mass of water lost}}{\text{Initial mass of cheese}}\right) \times 100\]This formula is used to assess the water content in any similar sample, serving as a reliable approach to measure and understand product composition.
Water Content Determination
Water content determination is a pivotal analysis to evaluate the amount of water in various products, crucial in many industries. It not only affects the material's texture and flavor but also influences its stability and shelf life.
In the example of cheese, determining water content is vital for maintaining its quality and ensuring that it meets regulatory standards. By analyzing water content, manufacturers can:
  • Control the product's quality and ensure uniformity across batches.
  • Adjust the production process to meet desired moisture levels.
  • Predict the storage conditions and the shelf life of the product.
Each step of water content determination, from initial mass calculation to finding the proportion of water lost, provides valuable insights into product handling and improvement strategies, ensuring the final product meets the expected standards.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of an impure iron ore is approximately \(55 \% \mathrm{w} / \mathrm{w} \mathrm{Fe}\). If the amount of Fe in the sample is determined gravimetrically by isolating it as \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) what mass of sample is needed to ensure that we isolate at least \(1.0 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3} ?\)

Calcium is determined gravimetrically by precipitating \(\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) and isolating \(\mathrm{CaCO}_{3}\). After dissolving a sample in \(10 \mathrm{~mL}\) of water and \(15 \mathrm{~mL}\) of \(6 \mathrm{M} \mathrm{HCl}\), the resulting solution is heated to boiling and a warm solution of excess ammonium oxalate is added. The solution is maintained at \(80^{\circ} \mathrm{C}\) and \(6 \mathrm{M} \mathrm{NH}_{3}\) is added dropwise, with stirring, until the solution is faintly alkaline. The resulting precipitate and solution are removed from the heat and allowed to stand for at least one hour. After testing the solution for completeness of precipitation, the sample is filtered, rinsed with \(0.1 \% \mathrm{w} / \mathrm{v}\) ammonium oxalate, and dried for one hour at \(100-120^{\circ} \mathrm{C}\). The precipitate is transferred to a muffle furnace where it is converted to \(\mathrm{CaCO}_{3}\) by drying at \(500 \pm 25^{\circ} \mathrm{C}\) until constant weight. (a) Why is the precipitate of \(\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) converted to \(\mathrm{CaCO}_{3} ?\) (b) In the final step, if the sample is heated at too high of a temperature some \(\mathrm{CaCO}_{3}\) is converted to \(\mathrm{CaO}\). What effect would this have on the reported \(\% \mathrm{w} / \mathrm{w}\) Ca? (c) Why is the precipitant, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) added to a hot, acidic solution instead of a cold, alkaline solution?

A \(0.8612-\mathrm{g}\) sample of a mixture of \(\mathrm{NaBr}\), \(\mathrm{NaI}\), and \(\mathrm{NaNO}_{3}\) is analyzed by adding \(\mathrm{AgNO}_{3}\) and precipitating a \(1.0186-\mathrm{g}\) mixture of \(\mathrm{AgBr}\) and AgI. The precipitate is then heated in a stream of \(\mathrm{Cl}_{2}\), which converts it to \(0.7125 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the \(\% \mathrm{w} / \mathrm{w} \mathrm{NaNO}_{3}\) in the sample.

In the presence of water vapor the surface of zirconia, \(\mathrm{ZrO}_{2}\), chemically adsorbs \(\mathrm{H}_{2} \mathrm{O},\) forming surface hydroxyls, \(\mathrm{ZrOH}\) (additional water is physically adsorbed as \(\mathrm{H}_{2} \mathrm{O}\) ). When heated above \(200^{\circ} \mathrm{C}\), the surface hydroxyls convert to \(\mathrm{H}_{2} \mathrm{O}(g),\) releasing one molecule of water for every two surface hydroxyls. Below \(200^{\circ} \mathrm{C}\) only physically absorbed water is lost. Nawrocki, et al. used thermogravimetry to determine the density of surface hydroxyls on a sample of zirconia that was heated to \(700^{\circ} \mathrm{C}\) and cooled in a desiccator containing humid \(\mathrm{N}_{2}{ }^{15}\) Heating the sample from \(200^{\circ} \mathrm{C}\) to \(900^{\circ} \mathrm{C}\) released \(0.006 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) for every gram of dehy- droxylated \(\mathrm{ZrO}_{2}\). Given that the zirconia had a surface area of \(33 \mathrm{~m}^{2} / \mathrm{g}\) and that one molecule of \(\mathrm{H}_{2} \mathrm{O}\) forms two surface hydroxyls, calculate the density of surface hydroxyls in \(\mu \mathrm{mol} / \mathrm{m}^{2}\).

Asolidsamplehasapproximately equalamounts of two or more of the following soluble salts: \(\mathrm{AgNO}_{3}, \mathrm{ZnCl}_{2}, \mathrm{~K}_{2} \mathrm{CO}_{3}, \mathrm{MgSO}_{4}, \mathrm{Ba}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\), and \(\mathrm{NH}_{4} \mathrm{NO}_{3} .\) A sample of the solid, sufficient to give at least 0.04 moles of any single salt, is added to \(100 \mathrm{~mL}\) of water, yielding a white precipitate and a clear solution. The precipitate is collected and rinsed with water. When a portion of the precipitate is placed in dilute \(\mathrm{HNO}_{3}\) it completely dissolves, leaving a colorless solution. A second portion of the precipitate is placed in dilute \(\mathrm{HCl}\), yielding a solid and a clear solution; when its filtrate is treated with excess \(\mathrm{NH}_{3}\), a white precipitate forms. Identify the salts that must be present in the sample, the salts that must be absent, and the salts for which there is insufficient information to make this determination. \({ }^{13}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free