Answer to Question #87483 in Organic Chemistry for Diana

Question

What is the mass (in g) of element in 1 mole of acesulfame potassium as well as the percentage?

Answer 1

The molecular formula of acesulfame potassium is C₄H₄KNO₄S.

M_ac = 12×4 + 1×4 + 39 +14 + 16×4 + 32 = 201 g/mol.

If n_ac = 1mol, m_ac = 201 g.

m(C) = A_r(C)×n(C) = 12g/mol×4mol = 48 g.

w(C) = m(C)/m_ac × 100% =48g/201g × 100% = 23.88%.

m(H) = A_r(H)×n(H) = 1g/mol×4mol = 4 g.

w(H) = m(H)/m_ac × 100% = 4g/201g × 100% = 1.99%.

m(K) = A_r(K)×n(K) = 39g/mol×1mol = 39 g.

w(K) = m(K)/m_ac × 100% = 39g/201g × 100% = 19.40%.

m(N) = A_r(N)×n(N) = 14g/mol×1mol = 14 g.

w(N) = m(N)/m_ac × 100% = 14g/201g × 100% = 6.97%.

m(O) = A_r(O)×n(O) = 16g/mol×4mol = 64 g.

w(O) = m(O)/m_ac × 100% = 64g/201g × 100% = 31.84%.

m(S) = A_r(S)×n(S) =32g/mol×1mol = 32 g.

w(S) = m(S)/m_ac × 100% = 32g/201g × 100% = 15.92%.