In the example part 4.27g of carbon and 5.69g of oxygen was written...so how come in the solution I saw 1.33 g of oxygen n and 1g of carbon 

We all knew that chemicals law explain the behaviors of chemicals which hncludes, laws of: cnnservation of mass or matter, definite propoption or constant compositions and multiple proportion. Statement that, mass or matter can neither bf created nor destroyed in a curse of a chemical reaction, all pure chemical compounds contains thd same compound combine in the proportion by mass and if two elements combine together to form more than one compound, several masses of A are separately fixed with mass B are in a whole number ratio respectively.

<h2>Testing Dalton’s Atomic Theory</h2>
<p>In the following drawing, the green spheres represent atoms of a certain element. The blue spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these <a data-lesson="revising-chemical-symbols" data-subject="chemistry" data-tutorial="physical-and-chemical-change" href="https://app.ulearngo.com/chemistry/physical-and-chemical-change/revising-chemical-symbols">symbols</a> violate any of the ideas of Dalton’s atomic theory? If so, which one?</p>
<div class="wp-caption alignnone" id="attachment_96165" style="width: 650px"><a href="https://data.ulearngo.com/assets/94b0f8e7-f824-4616-b045-27384fbb2abd"><img alt="dalton-test" class="size-full wp-image-96165" height="106" src="https://data.ulearngo.com/assets/94b0f8e7-f824-4616-b045-27384fbb2abd" width="650"/></a><p class="wp-caption-text">Image credit: OpenStax, Chemistry</p></div>
<h3>Solution</h3>
<p>The starting <a data-lesson="materials" data-subject="chemistry" data-tutorial="classification-of-matter" href="https://app.ulearngo.com/chemistry/classification-of-matter/materials">materials</a> consist of two green spheres and two blue spheres. The products consist of only one green sphere and one blue sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)</p>
<h2>Law of Definite Proportions</h2>
<p>Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that <em data-effect="italics">all samples of a pure compound contain the same <a data-lesson="elements" data-subject="chemistry" data-tutorial="classification-of-matter" href="https://app.ulearngo.com/chemistry/classification-of-matter/elements">elements</a> in the same proportion by mass</em>. This statement is known as the <strong>law of definite proportions</strong> or the <strong>law of constant composition</strong>.</p>
<p>The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of <a data-lesson="enthalpy-of-combustion" data-subject="chemistry" data-tutorial="thermochemistry" href="https://app.ulearngo.com/chemistry/thermochemistry/enthalpy-of-combustion">isooctane</a> (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33 : 1, as shown in the table below.</p>
<table class="span-all" id="fs-idp114534448" summary="This table displays how much carbon and hydrogen are in samples A, B and C, as well as the mass ratio of each. Sample A contains 14.82 grams of carbon and 2.78 grams of hydrogen. Its mass ratio is, therefore, 14.82 grams of carbon over 2.78 grams of hydrogen. This ratio is equal to 5.33 grams of carbon over 1.00 gram of hydrogen. Sample B contains 22.33 grams of carbon and 4.19 grams of hydrogen. Its mass ratio is 22.3 grams of carbon over 4.19 grams of hydrogen which is equal to 5.33 grams of carbon over 1.00 gram of hydrogen. Sample C contains 19.40 grams of carbon and 3.64 grams of hydrogen. Its mass ratio is 19.40 grams of carbon over 3.63 grams of hydrogen which is equal to 5.33 grams of carbon over 1.00 gram of hydrogen.">
<thead>
<tr>
<th colspan="4" data-align="center" scope="col">
<h3>Constant Composition of Isooctane</h3>
</th>
</tr>
<tr valign="top">
<th data-align="left" scope="col">Sample</th>
<th data-align="left" scope="col">Carbon</th>
<th data-align="left" scope="col">Hydrogen</th>
<th data-align="left" scope="col">Mass Ratio</th>
</tr>
</thead>
<tbody>
<tr valign="top">
<td data-align="center">A</td>
<td data-align="center">14.82 g</td>
<td data-align="center">2.78 g</td>
<td data-align="center">\(\frac{14.82\text{ g carbon}}{2.78\text{ g hydrogen}} = \frac{5.33\text{ g carbon}}{1.00\text{ g hydrogen}}\)</td>
</tr>
<tr valign="top">
<td data-align="center">B</td>
<td data-align="center">22.33 g</td>
<td data-align="center">4.19 g</td>
<td data-align="center">\(\frac{22.33\text{ g carbon}}{4.19\text{ g hydrogen}} = \frac{5.33\text{ g carbon}}{1.00\text{ g hydrogen}}\)</td>
</tr>
<tr valign="top">
<td data-align="center">C</td>
<td data-align="center">19.40 g</td>
<td data-align="center">3.64 g</td>
<td data-align="center">\(\frac{19.40\text{ g carbon}}{3.64\text{ g hydrogen}} = \frac{5.33\text{ g carbon}}{1.00\text{ g hydrogen}}\)</td>
</tr>
</tbody>
</table>
<p>It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many <a data-lesson="compounds-2" data-subject="chemistry" data-tutorial="quantitative-aspects-chemical-change" href="https://app.ulearngo.com/chemistry/quantitative-aspects-chemical-change/compounds-2">compounds</a> other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33 : 1.00.</p>
<h2>Law of Multiple Proportions</h2>
<p>Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The <strong>law of multiple proportions</strong> states that <em data-effect="italics">when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole numbers</em>.</p>
<p>For example, copper and chlorine can form a green, crystalline <a data-lesson="melting-and-freezing" data-subject="chemistry" data-tutorial="liquids-and-solids" href="https://app.ulearngo.com/chemistry/liquids-and-solids/melting-and-freezing">solid</a> with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio.</p>
<p>\(\cfrac{\frac{1.116\text{ g Cl}}{1\text{ g Cu}}}{\frac{0.558\text{ g Cl}}{1\text{ g Cu}}} = \cfrac{2}{1}\)</p>
<p id="fs-idp35032848">This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.</p>
<p id="fs-idm113441696">This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (<a href="#attachment_96169">see image below</a>).</p>
<div class="wp-caption alignnone" id="attachment_96169" style="width: 800px"><a href="https://data.ulearngo.com/assets/2d5f093b-2063-4283-bdf5-ee000d7a18b3"><img alt="multiple-proportions" class="size-full wp-image-96169" height="299" src="https://data.ulearngo.com/assets/2d5f093b-2063-4283-bdf5-ee000d7a18b3" width="800"/></a><p class="wp-caption-text">Compared to the copper chlorine compound in (a), where copper is represented by brown spheres and chlorine by green spheres, the copper chlorine compound in (b) has twice as many chlorine atoms per copper atom. Image credit:- OpenStax, Chemistry; a: modification of work by “Benjah-bmm27”/Wikimedia Commons; b: modification of work by “Walkerma”/Wikimedia Commons</p></div>
<h2>Laws of Definite and Multiple Proportions: Example</h2>
<p>A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?</p>
<h3>Solution</h3>
<p>In compound A, the mass ratio of carbon to oxygen is:</p>
<p>\(\cfrac{1.33\text{ g O}}{1\text{ g C}}\)</p>
<p>In compound B, the mass ratio of carbon to oxygen is:</p>
<p>\(\cfrac{2.67\text{ g O}}{1\text{ g C}}\)</p>
<p>The ratio of these ratios is:</p>
<p>\(\cfrac{\frac{1.33\text{ g O}}{1\text{ g C}}}{\frac{2.67\text{ g O}}{1\text{ g C}}} = \cfrac{1}{2}\)</p>
<p>This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much carbon per amount of oxygen (or twice as much oxygen per amount of carbon) as B. A possible pair of compounds that would fit this relationship would be A = CO<sub>2</sub> and B = CO.</p>

Chemistry generally explores the substances of which matter is composed, the investigation of their properties and reactions, and the use of such reactions to form new substances. 

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Dalton’s Atomic Theory Continued

Discover the composition of matter, early ideas in atomic theory, atomic structure and symbolism, the periodic table, and chemical nomenclature in this tutorial on atoms, molecules, and ions.


Dalton’s Atomic Theory Continued

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