{"id":27288,"date":"2025-06-19T22:00:20","date_gmt":"2025-06-19T22:00:20","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=27288"},"modified":"2025-06-19T22:00:23","modified_gmt":"2025-06-19T22:00:23","slug":"a-0-750-g-sample-of-tin-metal-reacts-with-0-201-g-of-oxygen-gas-to-form-tin-oxide","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-0-750-g-sample-of-tin-metal-reacts-with-0-201-g-of-oxygen-gas-to-form-tin-oxide\/","title":{"rendered":"A 0.750 g sample of tin metal reacts with 0.201 g of oxygen gas to form tin oxide."},"content":{"rendered":"\n

A 0.750 g sample of tin metal reacts with 0.201 g of oxygen gas to form tin oxide. Calculate the empirical formula of the tin oxide.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the empirical formula<\/strong> of the tin oxide, we need to determine the simplest whole-number ratio<\/strong> of atoms of tin (Sn) to atoms of oxygen (O) in the compound.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Convert mass to moles<\/h3>\n\n\n\n

Molar mass of Sn (tin)<\/strong> = 118.71 g\/mol
Molar mass of O (oxygen atom)<\/strong> = 16.00 g\/mol<\/p>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • Mass of Sn = 0.750 g<\/li>\n\n\n\n
  • Mass of O = 0.201 g<\/li>\n<\/ul>\n\n\n\n

    Now convert these masses to moles:Moles of Sn=0.750 g118.71 g\/mol\u22480.00632 mol\\text{Moles of Sn} = \\frac{0.750\\ \\text{g}}{118.71\\ \\text{g\/mol}} \\approx 0.00632\\ \\text{mol}Moles of Sn=118.71 g\/mol0.750 g\u200b\u22480.00632 molMoles of O=0.201 g16.00 g\/mol=0.01256 mol\\text{Moles of O} = \\frac{0.201\\ \\text{g}}{16.00\\ \\text{g\/mol}} = 0.01256\\ \\text{mol}Moles of O=16.00 g\/mol0.201 g\u200b=0.01256 mol<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Determine the mole ratio<\/h3>\n\n\n\n

    Divide each mole value by the smaller of the two:0.006320.00632=1.00\\frac{0.00632}{0.00632} = 1.000.006320.00632\u200b=1.000.012560.00632\u22481.99\u22482\\frac{0.01256}{0.00632} \\approx 1.99 \\approx 20.006320.01256\u200b\u22481.99\u22482<\/p>\n\n\n\n

    So, the mole ratio of Sn to O is approximately 1:2<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Write the empirical formula<\/h3>\n\n\n\n

    The empirical formula is based on the smallest whole-number ratio of atoms:<\/p>\n\n\n\n

      \n
    • 1 atom of tin (Sn)<\/li>\n\n\n\n
    • 2 atoms of oxygen (O)<\/li>\n<\/ul>\n\n\n\n

      Empirical formula=SnO2\\text{Empirical formula} = \\text{SnO}_2Empirical formula=SnO2\u200b<\/p>\n\n\n\n

      \n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      SnO\u2082<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      Explanation:<\/h3>\n\n\n\n

      An empirical formula shows the simplest ratio of elements in a compound. To find it, we convert the given masses of the elements to moles. Using the molar masses, we find that there are about 0.00632 mol of tin and 0.01256 mol of oxygen. We then divide both values by the smaller number of moles to simplify the ratio. The resulting 1:2 ratio means that for every 1 atom of tin, there are 2 atoms of oxygen in the compound. Therefore, the empirical formula is SnO\u2082.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      A 0.750 g sample of tin metal reacts with 0.201 g of oxygen gas to form tin oxide. Calculate the empirical formula of the tin oxide. The Correct Answer and Explanation is: To find the empirical formula of the tin oxide, we need to determine the simplest whole-number ratio of atoms of tin (Sn) to […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-27288","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27288","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=27288"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27288\/revisions"}],"predecessor-version":[{"id":27296,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/27288\/revisions\/27296"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=27288"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=27288"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=27288"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}