Two samples of sodium chloride are decomposed into their constituent elements. One sample produces \(6.98 \mathrm{~g}\) of sodium and \(10.7 \mathrm{~g}\) of chlorine, and the other sample produces \(11.2 \mathrm{~g}\) of sodium and \(17.3 \mathrm{~g}\) of chlorine. Are these results consistent with the law of definite proportions? Explain your answer.

Short Answer

Expert verified
Mass ratio for Sample 1 = 1.5344 g/g, and for Sample 2 = 1.5446 g/g. Since the ratios are very close, the results are consistent with the law of definite proportions.

Step by step solution

01

Calculate the Mass Ratios

To determine if the law of definite proportions holds, you need to calculate the mass ratio of chlorine to sodium for both samples. For the first sample, the mass ratio is calculated by dividing the mass of chlorine by the mass of sodium. Use the formula: mass ratio of Sample 1 = mass of chlorine / mass of sodium. Perform the calculation: mass ratio of Sample 1 = 10.7 g / 6.98 g.
02

Repeat the Calculation for the Second Sample

Calculate the same mass ratio for the second sample of sodium chloride using the analogous formula: mass ratio of Sample 2 = mass of chlorine / mass of sodium. Now perform the calculation: mass ratio of Sample 2 = 17.3 g / 11.2 g.
03

Analyze the Results

Compare the mass ratios obtained in the previous steps. If the ratios are the same (or very close allowing for experimental error), this supports the law of definite proportions, which states that a chemical compound always contains exactly the same proportion of elements by mass. Calculate and compare the decimal mass ratios to determine if they are consistent.

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Key Concepts

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

Mass Ratio in Chemistry
The concept of mass ratio is fundamental to understanding chemical reactions. It refers to the ratio of the masses of substances involved in the reaction. Specifically, in the context of a compound like sodium chloride (NaCl), the mass ratio is the weight of chlorine divided by the weight of sodium in a given sample.

This ratio is essential because it demonstrates the law of definite proportions, which posits that a chemical compound will always be composed of the same elements in the same proportion by mass, regardless of the size or source of the sample. For example, if two different samples of NaCl are broken down into sodium and chlorine, the mass ratios of chlorine to sodium should be the same for both samples, assuming there are no experimental errors. It is this consistency that was tested in the original exercise by comparing the mass ratios from two disparate samples.
Chemical Compound Composition
Understanding the composition of a chemical compound is crucial in chemistry. Each compound has a unique formula that specifies the types and numbers of atoms present. In the case of sodium chloride, for every atom of sodium, there is one atom of chlorine. This 1:1 ratio remains constant, which is an example of the law of definite proportions at the atomic level.

The composition of compounds is not only described in terms of atoms but also in terms of mass, as seen in the exercise. When you break down a compound, the mass of each element derived from it is not arbitrary but is an intrinsic property of the compound. This concept of composition is intertwined with the mass ratio, as changes in composition logically affect the mass ratio of the elements that make up the compound.
Stoichiometry
Stoichiometry is the section of chemistry that deals with the quantification of reactants and products in a chemical reaction. It allows chemists to predict the amounts of products that will form in a given reaction or the amounts of reactants needed to create a certain amount of product. The stoichiometric coefficients indicate the ratios of reactants and products involved in the reaction.

For instance, when decomposing sodium chloride, stoichiometry tells us that one mole of NaCl will produce one mole of sodium and one mole of chlorine gas, based on the balanced chemical equation. This mole ratio translates to a fixed mass ratio because each element has a specific molar mass, which is another way to express the law of definite proportions. When there's a deviation from the stoichiometric ratios, it raises a question about the purity of the reactants or the occurrence of experimental errors.
Experimental Error in Chemistry
Experimental error is a natural part of any scientific investigation, encompassing all the uncertainties and deviations that can occur when measuring, sampling, or experimenting. These errors can be systematic, resulting from flawed equipment or procedures, or random, as a result of unpredictable variations in observations.

In our exercise, if the mass ratios of chlorine to sodium from two samples of sodium chloride differ, it might initially suggest a contradiction to the law of definite proportions. However, it's important to consider experimental error as a possible explanation for any discrepancy. Small variations in measurements are expected, and only significant differences would challenge the law. Accurate analytical techniques are crucial to minimize these errors and confirm the consistency required by chemical laws.

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Most popular questions from this chapter

Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these are decomposed, the sulfur dioxide produces \(3.49 \mathrm{~g}\) oxygen and \(3.50 \mathrm{~g}\) sulfur, while the sulfur trioxide produces \(6.75 \mathrm{~g}\) oxygen and \(4.50 \mathrm{~g}\) sulfur. Calculate the mass of oxygen per gram of sulfur for each sample and show that these results are consistent with the law of multiple proportions.

Discuss these questions with the group and record your consensus answer. The table shown here includes data similar to those used by Mendeleev when he created the periodic table. On a small card, write the symbol, atomic mass, and a stable compound formed by each element. Without consulting a periodic table, arrange the cards so that atomic mass increases from left to right and elements with similar properties are above and below each other. Copy the periodic table you have invented onto a piece of paper. There is one element missing. Predict its mass and a stable compound it might form. $$ \begin{array}{ccc} \text { Element } & \text { Atomic Mass } & \text { Stable Compound } \\ \hline \text { Be } & 9 & \text { BeCI }_{2} \\ \hline \text { S } & 32 & \mathrm{H}_{2} \mathrm{~S} \\ \hline \mathrm{F} & 19 & \mathrm{~F}_{2} \\ \hline \mathrm{Ca} & 40 & \mathrm{CaCl}_{2} \\ \hline \mathrm{Li} & 7 & \mathrm{LiCl} \\ \hline \mathrm{Si} & 28 & \mathrm{SiH}_{4} \\ \hline \mathrm{Cl} & 35.4 & \mathrm{Cl}_{2} \\ \hline \mathrm{B} & 10.8 & \mathrm{BH}_{3} \\ \hline \mathrm{Ge} & 72.6 & \mathrm{GeH}_{4} \\ \hline \mathrm{N} & 14 & \mathrm{NF}_{3} \\ \hline \mathrm{O} & 16 & \mathrm{H}_{2} \mathrm{O} \\ \hline \mathrm{Ga} & 69.7 & \mathrm{GaH}_{3} \\ \hline \hline \text { As } & 75 & \text { AsF }_{3} \\ \hline \text { C } & 12 & \text { CH }_{4} \\ \hline \text { K } & 39 & \text { KCl } \\ \hline \text { Mg } & 24.3 & \text { MgCl }_{2} \\ \hline \text { Se } & 79 & \text { H }_{2} \text { Se } \\ \hline \text { Al } & 27 & \text { AlH }_{3} \\ \hline \text { Br } & 80 & \text { Br }_{2} \\ \hline \text { Na } & 23 & \text { NaCl } \\ \hline \end{array} $$

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