Chemistry Labs

Problem 1

Gold nanoparticles modified with two kinds of single-stranded nucleic acid fragments a and b aggregate in the presence of the target nucleic acid a'b' from the coronavirus, causing a colour change of the solution from red to blue within minutes. Gold nanoparticles are composed of closely packed gold atoms with the density of solid gold, ρ=19.3\rho = 19.3 g cm−3^{-3}. (a) Calculate the number of gold atoms NN in a spherical gold nanoparticle of diameter 30.0 nm (M(Au)=197.0M(\ce{Au}) = 197.0 g mol−1^{-1}, NA=6.022×1023N_A = 6.022\times 10^{23} mol−1^{-1}). (b) 5.2 mg of HAuClX4 ⋅ 3 HX2O\ce{HAuCl4*3H2O} (M=394M = 394 g mol−1^{-1}) was completely converted into uniform spherical gold nanoparticles of diameter 30.0 nm in 100.0 mL of solution; the absorbance of the red solution measured at 530 nm in a 1 cm cuvette was 0.800. Calculate the molar extinction coefficient (per mole of nanoparticles) at 530 nm.
Step 2 of 3: Nanoparticle concentration
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cNP=n(Au)N V=1.32×10−58.3×105×0.1000=1.6×10−7 mol L−1c_{\text{NP}} = \dfrac{n(\ce{Au})}{N\,V} = \dfrac{1.32\times 10^{-5}}{8.3\times 10^{5}\times 0.1000} = 1.6\times 10^{-7}\ \text{mol L}^{-1}
Analysis

The precursor supplies 5.2×10−3/394=1.32×10−55.2\times 10^{-3}/394 = 1.32\times 10^{-5} mol of Au atoms. One mole of nanoparticles contains 8.3×1058.3\times 10^{5} mol of Au, so nNP=1.32×10−5/8.3×105=1.6×10−11n_{\text{NP}} = 1.32\times 10^{-5}/8.3\times 10^{5} = 1.6\times 10^{-11} mol in 100.0 mL, i.e. cNP=1.6×10−7c_{\text{NP}} = 1.6\times 10^{-7} mol L−1^{-1}.