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class="mark-page-favorite pull-right" data-pid="722" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../343/stoichiometry.html">Stoichiometry</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">The mole concept</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 45 minutes"><i class="fa fa-clock-o"></i> 45'</span> </ol> <article id="main-article"> <p><img alt="" src="../../images/test-images/themoletoo-1.png" style="float: left; width: 160px; height: 92px;">A mole is simply a number, specifically Avogadro's number (<em>L</em>) which is 6.02 ×10<sup>23</sup>. It is the number of atoms in exactly 12 g of pure carbon-12.</p> <p>Mole calculations are not difficult in terms of the mathematical functions that you need to carry out, but it can be difficult to work out <em>what</em> needs to be added, divided, multiplied etc<strong>. </strong> <strong>Use revision card 4 to review how the the empirical formula of a compound can be experimentally determined</strong>. Then <strong>use the practice questions to ensure you are confident calculating the formulae from raw data</strong>.</p> <hr class="hidden-separator"> <div class="panel panel-has-colored-body panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <p>Ensure you are confident using the terms below and learn the asterisked* definitions</p> <p>Avogadro constant (<em>L</em>)*, molar mass (<em>M</em>), mole (<em>n</em>)*, empirical formula, molecular formula, percentage by mass</p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1" data-mode="" data-topics="589" data-subject-id="7" data-n-flashcards="5" style="text-align:center">Show flashcards</a></div><hr> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>What is a mole?</p> </div> </div> <div class="panel-body"> <div> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/351482553"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Using moles: Finding empirical and molecular formulae</p> </div> </div> <div class="panel-body"> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/378281924"></iframe></div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <p> The revision cards contain all of the essential content:</p> <div id="carousel-138" class="dynamic-gallery carousel slide" data-id="138"><div class="carousel-inner" role="listbox"><div class="item active"><a class="fancy" href="../../../std-galleries/7-138/screenshot-2019-12-24-at-121051.png" data-fancybox="gallery-138" title="" data-caption=""><img alt="" src="../../../std-galleries/7-138/screenshot-2019-12-24-at-121051.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-138/screenshot-2019-09-05-at-205801.png" data-fancybox="gallery-138" title="" data-caption=""><img alt="" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-205801.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-138/screenshot-2019-09-05-at-204827.png" data-fancybox="gallery-138" title="" data-caption=""><img alt="" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204827.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-138/screenshot-2019-09-05-at-204847.png" data-fancybox="gallery-138" title="" data-caption=""><img alt="" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204847.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-138/screenshot-2019-09-05-at-204904.png" data-fancybox="gallery-138" title="" data-caption=""><img alt="" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204904.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-138/screenshot-2019-09-05-at-204922.png" data-fancybox="gallery-138" title="" data-caption=""><img alt="" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204922.png"></a></div></div><a class="left carousel-control" href="#carousel-138" role="button" data-slide="prev"><i class="fa fa-fw fa-chevron-left"></i></a><a class="right carousel-control" href="#carousel-138" role="button" data-slide="next"><i class="fa fa-fw fa-chevron-right"></i></a></div><ol class="std-carousel-indicators"><li data-index="0"><img title="Click to view" src="../../../std-galleries/7-138/screenshot-2019-12-24-at-121051-thumb128.jpg"><li><li data-index="1"><img title="Click to view" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-205801-thumb128.jpg"><li><li data-index="2"><img title="Click to view" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204827-thumb128.jpg"><li><li data-index="3"><img title="Click to view" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204847-thumb128.jpg"><li><li data-index="4"><img title="Click to view" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204904-thumb128.jpg"><li><li data-index="5"><img title="Click to view" src="../../../std-galleries/7-138/screenshot-2019-09-05-at-204922-thumb128.jpg"><li></li></ol> <p> </p> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test yourself</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <div class="tib-quiz" data-stats="7-148-722"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the relative molecular mass (M<sub>r</sub>) of magnesium nitrate, Mg(NO<sub>3</sub>)<sub>2</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>86.32</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>148.33</span></label> </p><p><label class="radio"> <input type="radio"> <span>148</span></label> </p><p><label class="radio"> <input type="radio"> <span>86</span></label> </p></div><div class="q-explanation"><p>To find the M<sub>r</sub> the relative atomic masses of all the atoms are added together. Remember to use the values from the data book, given to <strong>two decimal places</strong>:</p><p>24.31+((14.01+16.00×3)×2) = 24.31+124.02 = 148.33</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the molar mass of carbon, and its units?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>12 g mol<sup>−1</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>12.01 g mol<sup>−1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>12.01 (no units)</span></label> </p><p><label class="radio"> <input type="radio"> <span>12 (no units)</span></label> </p></div><div class="q-explanation"><p>The molar mass is the <strong>mass of one mole</strong>, and therefore has units of grams per mole (g mol<sup>−1 </sup>or g/mol).</p><p>Relative atomic/molecular/formula masses have no units (as they are relative) but <strong>molar mass</strong> does have units.</p><p>Remember to use the relative atomic mass values from the data book, given to <strong>two decimal places</strong> when calculating the molar mass.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>How many atoms of hydrogen are present in one mole of ethane, C<sub>2</sub>H<sub>6</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>3.61×10<sup>23</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>36×10<sup>24</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>6</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>3.61×10<sup>24</sup></span></label> </p></div><div class="q-explanation"><p>One mole of ethane is 6.02×10<sup>23</sup> molecules of ethane (<em>L</em>, Avogadro's constant = 6.02×10<sup>23</sup>).</p><p>Each molecule has six hydrogen atoms in it. So there will be 6 × 6.02×10<sup>23</sup> atoms of hygrogen.</p><p>6 × 6.02×10<sup>23</sup> = 36.12×10<sup>23</sup> = 3.61×10<sup>24</sup> given to 3 sig figs.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>How many chloride ions will be released into water when 0.250 mol of calcium chloride (CaCl<sub>2</sub>) is dissolved completely in water?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>0.500</span></label> </p><p><label class="radio"> <input type="radio"> <span>6.02×10<sup>23</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>1.505×10<sup>23</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>3.01×10<sup>23</sup></span></label> </p></div><div class="q-explanation"><p>One mole of calcium chloride is 6.02×10<sup>23</sup> 'formula units' of CaCl<sub>2</sub> (<em>L</em>, Avogadro's constant = 6.02×10<sup>23</sup>).</p><p>Each 'formula unit' will release two chloride ions: CaCl<sub>2</sub> → Ca<sup>2+</sup> + 2Cl<sup>−</sup></p><p>So, 0.250 moles of calcium chloride, will release 0.500 moles of chloride ions:</p><p>0.500 × 6.02×10<sup>23</sup> = 3.01×10<sup>23</sup> given to 3 sig figs.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>To two significant figures<strong>, </strong>how many atoms of oxygen are present in 640g of oxygen?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>2.4×10<sup>25</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>40</span></label> </p><p><label class="radio"> <input type="radio"> <span>12×10<sup>24</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>1.2×10<sup>25</sup></span></label> </p></div><div class="q-explanation"><p>One mole of oxygen atoms is 6.02×10<sup>23</sup> atoms (<em>L</em>, Avogadro's constant = 6.02×10<sup>23</sup>).</p><p>Moles = mass/molar mass, so 640/16.00 = 40 mol</p><p>40 moles of oxygen atoms will be 40 × 6.02×10<sup>23</sup> atoms of oxygen.</p><p>40 × 6.02×10<sup>23</sup> = 240.8×10<sup>23</sup> = 2.408×10<sup>25</sup></p><p>Thus 2.4×10<sup>25</sup> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>A compound has an empirical formula of CH, and a molecular mass of approximately 104. What is the molecular formula of the compound?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>C<sub>15</sub>H<sub>15</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>C<sub>8</sub>H<sub>8</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>7</sub>H<sub>20</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>6</sub>H<sub>6</sub></span></label> </p></div><div class="q-explanation"><p>The empirical formula is the simplest ratio of atoms in the molecule/formula. The molecular formula represents the actual number of atoms present in the molecule.</p><p>The empirical 'formula unit' has a mass of approx. 13 (12+1), and the molecular mass of the molecule is 104.</p><p>104 / 13 = 8</p><p>So there are eight 'formula units' of the empirical formula in the molecule: The molecular formula is C<sub>8</sub>H<sub>8</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>A sample of a pure compound from a vanilla plant is analysed and shown to consist of 48.04g of carbon, 4.04g of hydrogen and 24.00g of oxygen by mass. What is the empirical formula of the compound?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>C<sub>4</sub>H<sub>4</sub>O<sub>2</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>8</sub>H<sub>4</sub>O<sub>3</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>C<sub>8</sub>H<sub>8</sub>O<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>3</sub>H<sub>3</sub>O<sub>4</sub></span></label> </p></div><div class="q-explanation"><p>Find the simplest ratio of the number of (moles of) atoms.</p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">Carbon</td><td style="text-align: center;">Hydrogen</td><td style="text-align: center;">Oxygen</td></tr><tr><td style="text-align: center;"><p>moles = mass/molar mass</p></td><td style="text-align: center;">48.04/12.01 = 4</td><td style="text-align: center;">4.04/1.01 = 4</td><td style="text-align: center;">24.00/16.00 = 1.5</td></tr><tr><td style="text-align: center;">multiply up to whole numbers (×2)</td><td style="text-align: center;">8</td><td style="text-align: center;">8</td><td style="text-align: center;">3</td></tr></tbody></table><p>Thus the answer is C<sub>8</sub>H<sub>8</sub>O<sub>3</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>A compound consists of 53% carbon, 6% hydrogen and 41% nitrogen by mass. What is the empirical formula of the compound?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>C<sub>6</sub>H<sub>4</sub>N<sub>4</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>C<sub>3</sub>H<sub>4</sub>N<sub>2</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>4</sub>H<sub>3</sub>N<sub>6</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>4</sub>H<sub>6</sub>N<sub>3</sub></span></label> </p></div><div class="q-explanation"><p>Find the simplest ratio of the number of (moles of) atoms.</p><p>Start by assuming 100g: Thus 53g of C atoms, 6g of H atoms, 41g of N atoms.</p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">Carbon</td><td style="text-align: center;">Hydrogen</td><td style="text-align: center;">Nitrogen</td></tr><tr><td style="text-align: center;"><p>moles = mass/molar mass</p></td><td style="text-align: center;">53/12.01 = 4.412989176</td><td style="text-align: center;">6/1.01 = 5.940594059</td><td style="text-align: center;">41/14.01 = 2.926481085</td></tr><tr><td style="text-align: center;"><p>divide through by smallest</p></td><td style="text-align: center;">4.41../2.92.. ≈ 1.5</td><td style="text-align: center;">5.94../2.92.. ≈ 2</td><td style="text-align: center;">2.92../2.92.. = 1</td></tr><tr><td style="text-align: center;">multiply up to whole numbers (×2)</td><td style="text-align: center;">3</td><td style="text-align: center;">4</td><td style="text-align: center;">2</td></tr></tbody></table><p>Thus the answer is C<sub>3</sub>H<sub>4</sub>N<sub>2</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>When phosphorus reacts with chlorine a yellow liquid is formed.</p><p>12.39g of phosphorus reacted completely to form 54.93g of the phosphorus chloride.</p><p>What is the empirical formula of the phosphorus chloride?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>PCl<sub>3</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>PCl<sub>2</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>PCl<sub>5</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>PCl<sub>4</sub></span></label> </p></div><div class="q-explanation"><p>Find the simplest ratio of the number of (moles of) atoms.</p><p>Start by finding the mass of chlorine that reacted:</p><p>Mass of chlorine = mass of phosphorus chloride minus mass of phosphorus = 54.93 − 12.39 = 42.54g</p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">Phosphorus</td><td style="text-align: center;">Chlorine</td></tr><tr><td style="text-align: center;"><p>moles = mass/molar mass</p></td><td style="text-align: center;">12.39/30.97 = 0.400064578</td><td style="text-align: center;">42.54/35.45 = 1.2</td></tr><tr><td style="text-align: center;"><p>divide through by smallest</p></td><td style="text-align: center;">0.400../0.400.. = 1</td><td style="text-align: center;">1.2 / 0.400.. ≈ 3</td></tr><tr><td style="text-align: center;">whole numbers</td><td style="text-align: center;">1</td><td style="text-align: center;">3</td></tr></tbody></table><p>Thus the answer is<sub> </sub>PCl<sub>3</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>A carbohydrate consists of 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass, and has a molar mass of approximately 120 g mol<sup>−1</sup>.</p><p>What is the molecular formula of the compound?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>C<sub>6</sub>H<sub>16</sub>O<sub>2</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>C<sub>4</sub>H<sub>2</sub>O<sub>4</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>C<sub>4</sub>H<sub>8</sub>O<sub>4</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>2</sub>O</span></label> </p></div><div class="q-explanation"><p>Find the simplest ratio of the number of (moles of) atoms.</p><p>Start by assuming 100g: Thus 40.0g of C atoms, 6.7g of H atoms, 53.3g of O atoms.</p><table border="0" cellpadding="0" cellspacing="0" style="width: 100%;"><tbody><tr><td style="text-align: center;"> </td><td style="text-align: center;">Carbon</td><td style="text-align: center;">Hydrogen</td><td style="text-align: center;">Oxygen</td></tr><tr><td style="text-align: center;"><p>moles = mass/molar mass</p></td><td style="text-align: center;">40.0/12.01 = 3.330557868</td><td style="text-align: center;">6.7/1.01 = 6.633663366</td><td style="text-align: center;">53.3/16.00 = 3.33125</td></tr><tr><td style="text-align: center;"><p>divide through by smallest</p></td><td style="text-align: center;">3.33../3.33.. = 1</td><td style="text-align: center;">6.63../3.33.. ≈ 2</td><td style="text-align: center;">3.33../3.33.. ≈ 1</td></tr><tr><td style="text-align: center;">whole numbers</td><td style="text-align: center;">1</td><td style="text-align: center;">2</td><td style="text-align: center;">1</td></tr></tbody></table><p>Thus the empirical formula is CH<sub>2</sub>O.</p><p>The empirical 'formula unit' has a mass of approx. 30 (12+2+16), and the molecular mass of the molecule is approx. 120.</p><p>120 / 30 = 4</p><p>So there are four 'formula units' of the empirical formula in the molecule: The molecular formula is<sub> </sub>C<sub>4</sub>H<sub>8</sub>O<sub>4</sub>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Exam-style questions</p> </div> </div> <div class="panel-body"> <div> <a href="../2123/stoichiometry-core-sl-hl-paper-1-questions.html" title="Stoichiometry core (SL & HL) paper 1 questions">Stoichiometry core (SL & HL) paper 1 questions</a></div> <div> </div> <div> <a href="../2353/stoichiometry-core-sl-hl-paper-2-questions.html" title="Stoichiometry core (SL & HL) paper 2 questions">Stoichiometry core (SL & HL) paper 2 questions</a></div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="page-container panel-self-assessment" data-id="722"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong>The mole concept</strong> have you understood?</p><div class="slider-container text-center"><div id="self-assessment-slider" class="sib-slider self-assessment " data-value="1" data-percentage=""></div></div> <label class="label-lg">My notes</label> <textarea name="page-notes" class="form-control" rows="3" placeholder="Write your notes here..."></textarea> </div> <div class="panel-footer text-xs-center"> 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