{"id":23795,"date":"2025-06-17T21:41:23","date_gmt":"2025-06-17T21:41:23","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=23795"},"modified":"2025-06-17T21:41:26","modified_gmt":"2025-06-17T21:41:26","slug":"calculate-the-mass-percent-of-nitrogen-in-leadii-nitrate","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-mass-percent-of-nitrogen-in-leadii-nitrate\/","title":{"rendered":"Calculate the mass percent of nitrogen in leadII) nitrate"},"content":{"rendered":"\n

Calculate the mass percent of nitrogen in leadII) nitrate. Round your answer to the nearest percentage.<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

To calculate the mass percent of nitrogen<\/strong> in lead(II) nitrate<\/strong>, we need to:<\/p>\n\n\n\n

\n\n\n\n

\u2705 Step 1: Write the chemical formula<\/strong><\/h3>\n\n\n\n

The formula for lead(II) nitrate<\/strong> is: Pb(NO3)2\\text{Pb(NO}_3)_2<\/p>\n\n\n\n

\n\n\n\n

\u2705 Step 2: Calculate the molar mass of Pb(NO\u2083)\u2082<\/strong><\/h3>\n\n\n\n
    \n
  • Atomic mass of Pb (lead)<\/strong> = 207.2 g\/mol<\/li>\n\n\n\n
  • Atomic mass of N (nitrogen)<\/strong> = 14.01 g\/mol<\/li>\n\n\n\n
  • Atomic mass of O (oxygen)<\/strong> = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Each nitrate ion (NO3)\u2212(\\text{NO}_3)^- has:<\/p>\n\n\n\n

      \n
    • 1 nitrogen<\/li>\n\n\n\n
    • 3 oxygens<\/li>\n<\/ul>\n\n\n\n

      Since there are two nitrate ions<\/strong> in the compound:<\/p>\n\n\n\n

        \n
      • Nitrogens: 2\u00d714.01=28.022 \\times 14.01 = 28.02 g\/mol<\/li>\n\n\n\n
      • Oxygens: 2\u00d73\u00d716.00=96.002 \\times 3 \\times 16.00 = 96.00 g\/mol<\/li>\n<\/ul>\n\n\n\n

        Now add all parts: Molar mass of Pb(NO3)2=207.2+28.02+96.00=331.22 g\/mol\\text{Molar mass of Pb(NO}_3)_2 = 207.2 + 28.02 + 96.00 = 331.22\\ \\text{g\/mol}<\/p>\n\n\n\n

        \n\n\n\n

        \u2705 Step 3: Calculate mass percent of nitrogen<\/strong><\/h3>\n\n\n\n

        Mass percent N=(Mass of N in 1 moleMolar mass of compound)\u00d7100\\text{Mass percent N} = \\left( \\frac{\\text{Mass of N in 1 mole}}{\\text{Molar mass of compound}} \\right) \\times 100 =(28.02331.22)\u00d7100\u22488.46%= \\left( \\frac{28.02}{331.22} \\right) \\times 100 \\approx 8.46\\%<\/p>\n\n\n\n

        Rounded to the nearest percent: 8%\\boxed{8\\%}<\/p>\n\n\n\n

        \n\n\n\n

        \u2705 Explanation (300 words)<\/strong><\/h3>\n\n\n\n

        To determine the mass percent of nitrogen in a compound like lead(II) nitrate, we must know both the molecular formula and the atomic masses of the constituent elements. Lead(II) nitrate has the formula Pb(NO\u2083)\u2082, indicating that it consists of one lead atom and two nitrate ions. Each nitrate ion contains one nitrogen atom and three oxygen atoms. Thus, the entire molecule includes one lead atom, two nitrogen atoms, and six oxygen atoms.<\/p>\n\n\n\n

        By referencing a periodic table, we find that the atomic masses are approximately: Pb = 207.2 g\/mol, N = 14.01 g\/mol, and O = 16.00 g\/mol. Calculating the mass contribution from each element and adding them gives the total molar mass of the compound. Nitrogen contributes 2\u00d714.01=28.022 \\times 14.01 = 28.02 g\/mol to the total mass of 331.22 g\/mol.<\/p>\n\n\n\n

        To find what percentage of the compound’s mass comes from nitrogen, divide the mass of nitrogen by the total molar mass and multiply by 100. This yields approximately 8.46%, which rounds to 8%<\/strong>.<\/p>\n\n\n\n

        Mass percent is useful in chemistry to understand the composition of a compound. It tells us how much of a particular element is present by weight in a given compound. This can be helpful in laboratory calculations, such as determining how much of a compound is needed to supply a specific amount of an element\u2014in this case, nitrogen.<\/p>\n","protected":false},"excerpt":{"rendered":"

        Calculate the mass percent of nitrogen in leadII) nitrate. Round your answer to the nearest percentage. The correct answer and explanation is: To calculate the mass percent of nitrogen in lead(II) nitrate, we need to: \u2705 Step 1: Write the chemical formula The formula for lead(II) nitrate is: Pb(NO3)2\\text{Pb(NO}_3)_2 \u2705 Step 2: Calculate the molar […]<\/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-23795","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23795","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=23795"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23795\/revisions"}],"predecessor-version":[{"id":23796,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23795\/revisions\/23796"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23795"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23795"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23795"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}